Curved antennas, such as the ones shown in Figure 1, are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function.
In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior.
The graph of a quadratic function is a U-shaped curve called a parabola . One important feature of the graph is that it has an extreme point, called the vertex . If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value . In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry . These features are illustrated in Figure 2.
The y-intercept is the point at which the parabola crosses the y-axis. The x-intercepts are the points at which the parabola crosses the x-axis. If they exist, the x-intercepts represent the zeros , or roots , of the quadratic function, the values of x x at which y = 0. y = 0.
Determine the vertex, axis of symmetry, zeros, and y - y - intercept of the parabola shown in Figure 3.
The vertex is the turning point of the graph. We can see that the vertex is at ( 3 , 1 ) . ( 3 , 1 ) . Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is x = 3. x = 3. This parabola does not cross the x - x - axis, so it has no zeros. It crosses the y - y - axis at ( 0 , 7 ) ( 0 , 7 ) so this is the y-intercept.
The general form of a quadratic function presents the function in the form
f ( x ) = a x 2 + b x + c f ( x ) = a x 2 + b x + cwhere a , b , a , b , and c c are real numbers and a ≠ 0. a ≠ 0. If a > 0 , a > 0 , the parabola opens upward. If a < 0 , a < 0 , the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry.
The axis of symmetry is defined by x = − b 2 a . x = − b 2 a . If we use the quadratic formula, x = − b ± b 2 − 4 a c 2 a , x = − b ± b 2 − 4 a c 2 a , to solve a x 2 + b x + c = 0 a x 2 + b x + c = 0 for the x - x - intercepts, or zeros, we find the value of x x halfway between them is always x = − b 2 a , x = − b 2 a , the equation for the axis of symmetry.
Figure 4 represents the graph of the quadratic function written in general form as y = x 2 + 4 x + 3. y = x 2 + 4 x + 3. In this form, a = 1 , b = 4 , a = 1 , b = 4 , and c = 3. c = 3. Because a > 0 , a > 0 , the parabola opens upward. The axis of symmetry is x = − 4 2 ( 1 ) = −2. x = − 4 2 ( 1 ) = −2. This also makes sense because we can see from the graph that the vertical line x = −2 x = −2 divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, ( −2 , −1 ) . ( −2 , −1 ) . The x - x - intercepts, those points where the parabola crosses the x - x - axis, occur at ( −3 , 0 ) ( −3 , 0 ) and ( −1 , 0 ) . ( −1 , 0 ) .
The standard form of a quadratic function presents the function in the form
f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + kwhere ( h , k ) ( h , k ) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function .
The standard form is useful for determining how the graph is transformed from the graph of y = x 2 . y = x 2 . Figure 6 is the graph of this basic function.
If k > 0 , k > 0 , the graph shifts upward, whereas if k < 0 , k < 0 , the graph shifts downward. In Figure 5, k > 0 , k > 0 , so the graph is shifted 4 units upward. If h > 0 , h > 0 , the graph shifts toward the right and if h < 0 , h < 0 , the graph shifts to the left. In Figure 5, h < 0 , h < 0 , so the graph is shifted 2 units to the left. The magnitude of a a indicates the stretch of the graph. If | a | >1 , | a | > 1 , the point associated with a particular x - x - value shifts farther from the x-axis, so the graph appears to become narrower, and there is a vertical stretch. But if | a | < 1 , | a | < 1 , the point associated with a particular x - x - value shifts closer to the x-axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure 5, | a | > 1 , | a | > 1 , so the graph becomes narrower.
The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.
a ( x − h ) 2 + k = a x 2 + b x + c a x 2 − 2 a h x + ( a h 2 + k ) = a x 2 + b x + c a ( x − h ) 2 + k = a x 2 + b x + c a x 2 − 2 a h x + ( a h 2 + k ) = a x 2 + b x + c
For the linear terms to be equal, the coefficients must be equal.
–2 a h = b , so h = − b 2 a –2 a h = b , so h = − b 2 aThis is the axis of symmetry we defined earlier. Setting the constant terms equal:
a h 2 + k = c k = c − a h 2 = c − a − ( b 2 a ) 2 = c − b 2 4 a a h 2 + k = c k = c − a h 2 = c − a − ( b 2 a ) 2 = c − b 2 4 a
In practice, though, it is usually easier to remember that k is the output value of the function when the input is h , h , so f ( h ) = k . f ( h ) = k .
A quadratic function is a polynomial function of degree two. The graph of a quadratic function is a parabola.
The general form of a quadratic function is f ( x ) = a x 2 + b x + c f ( x ) = a x 2 + b x + c where a , b , a , b , and c c are real numbers and a ≠ 0. a ≠ 0.
The standard form of a quadratic function is f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k where a ≠ 0. a ≠ 0.
The vertex ( h , k ) ( h , k ) is located at
h = – b 2 a , k = f ( h ) = f ( − b 2 a ) h = – b 2 a , k = f ( h ) = f ( − b 2 a )Given a graph of a quadratic function, write the equation of the function in general form.
Write an equation for the quadratic function g g in Figure 7 as a transformation of f ( x ) = x 2 , f ( x ) = x 2 , and then expand the formula, and simplify terms to write the equation in general form.
We can see the graph of g is the graph of f ( x ) = x 2 f ( x ) = x 2 shifted to the left 2 and down 3, giving a formula in the form g ( x ) = a ( x − ( −2 ) ) 2 − 3 = a ( x + 2 ) 2 – 3. g ( x ) = a ( x − ( −2 ) ) 2 − 3 = a ( x + 2 ) 2 – 3.
Substituting the coordinates of a point on the curve, such as ( 0 , −1 ) , ( 0 , −1 ) , we can solve for the stretch factor.
− 1 = a ( 0 + 2 ) 2 − 3 2 = 4 a a = 1 2 − 1 = a ( 0 + 2 ) 2 − 3 2 = 4 a a = 1 2In standard form, the algebraic model for this graph is ( g ) x = 1 2 ( x + 2 ) 2 – 3. ( g ) x = 1 2 ( x + 2 ) 2 – 3.
To write this in general polynomial form, we can expand the formula and simplify terms.
g ( x ) = 1 2 ( x + 2 ) 2 − 3 = 1 2 ( x + 2 ) ( x + 2 ) − 3 = 1 2 ( x 2 + 4 x + 4 ) − 3 = 1 2 x 2 + 2 x + 2 − 3 = 1 2 x 2 + 2 x − 1 g ( x ) = 1 2 ( x + 2 ) 2 − 3 = 1 2 ( x + 2 ) ( x + 2 ) − 3 = 1 2 ( x 2 + 4 x + 4 ) − 3 = 1 2 x 2 + 2 x + 2 − 3 = 1 2 x 2 + 2 x − 1
Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.
We can check our work using the table feature on a graphing utility. First enter Y1 = 1 2 ( x + 2 ) 2 − 3. Y1 = 1 2 ( x + 2 ) 2 − 3. Next, select TBLSET, TBLSET, then use TblStart = – 6 TblStart = – 6 and Δ Tbl = 2, Δ Tbl = 2, and select TABLE . TABLE . See Table 1.
| x x | –6 | –4 | –2 | 0 | 2 |
| y y | 5 | –1 | –3 | –1 | 5 |
The ordered pairs in the table correspond to points on the graph.
A coordinate grid has been superimposed over the quadratic path of a basketball in Figure 8. Assume that the point (–4, 7) is the highest point of the basketball’s trajectory. Find an equation for the path of the ball. Does the shooter make the basket?
Given a quadratic function in general form, find the vertex of the parabola.
Find the vertex of the quadratic function f ( x ) = 2 x 2 – 6 x + 7. f ( x ) = 2 x 2 – 6 x + 7. Rewrite the quadratic in standard form (vertex form).
The horizontal coordinate of the vertex will be at h = − b 2 a = − −6 2 ( 2 ) = 6 4 = 3 2 The vertical coordinate of the vertex will be at k = f ( h ) = f ( 3 2 ) = 2 ( 3 2 ) 2 − 6 ( 3 2 ) + 7 = 5 2 The horizontal coordinate of the vertex will be at h = − b 2 a = − −6 2 ( 2 ) = 6 4 = 3 2 The vertical coordinate of the vertex will be at k = f ( h ) = f ( 3 2 ) = 2 ( 3 2 ) 2 − 6 ( 3 2 ) + 7 = 5 2
Rewriting into standard form, the stretch factor will be the same as the a a in the original quadratic. First, find the horizontal coordinate of the vertex. Then find the vertical coordinate of the vertex. Substitute the values into standard form, using the " a a " from the general form.
f ( x ) = a x 2 + b x + c f ( x ) = 2 x 2 − 6 x + 7 f ( x ) = a x 2 + b x + c f ( x ) = 2 x 2 − 6 x + 7
The standard form of a quadratic function prior to writing the function then becomes the following:
f ( x ) = 2 ( x – 3 2 ) 2 + 5 2 f ( x ) = 2 ( x – 3 2 ) 2 + 5 2One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, k , k , and where it occurs, x . x .
Given the equation g ( x ) = 13 + x 2 − 6 x , g ( x ) = 13 + x 2 − 6 x , write the equation in general form and then in standard form.
Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y-values greater than or equal to the y-coordinate at the turning point or less than or equal to the y-coordinate at the turning point, depending on whether the parabola opens up or down.
The domain of any quadratic function is all real numbers unless the context of the function presents some restrictions.
The range of a quadratic function written in general form f ( x ) = a x 2 + b x + c f ( x ) = a x 2 + b x + c with a positive a a value is f ( x ) ≥ f ( − b 2 a ) , f ( x ) ≥ f ( − b 2 a ) , or [ f ( − b 2 a ) , ∞ ) ; [ f ( − b 2 a ) , ∞ ) ; the range of a quadratic function written in general form with a negative a a value is f ( x ) ≤ f ( − b 2 a ) , f ( x ) ≤ f ( − b 2 a ) , or ( − ∞ , f ( − b 2 a ) ] . ( − ∞ , f ( − b 2 a ) ] .
The range of a quadratic function written in standard form f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k with a positive a a value is f ( x ) ≥ k ; f ( x ) ≥ k ; the range of a quadratic function written in standard form with a negative a a value is f ( x ) ≤ k . f ( x ) ≤ k .
Given a quadratic function, find the domain and range.
Find the domain and range of f ( x ) = − 5 x 2 + 9 x − 1. f ( x ) = − 5 x 2 + 9 x − 1.
As with any quadratic function, the domain is all real numbers.
Because a a is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x - x - value of the vertex.
h = − b 2 a = − 9 2 ( −5 ) = 9 10 h = − b 2 a = − 9 2 ( −5 ) = 9 10The maximum value is given by f ( h ) . f ( h ) .
f ( 9 10 ) = -5 ( 9 10 ) 2 + 9 ( 9 10 ) − 1 = 61 20 f ( 9 10 ) = -5 ( 9 10 ) 2 + 9 ( 9 10 ) − 1 = 61 20
The range is f ( x ) ≤ 61 20 , f ( x ) ≤ 61 20 , or ( − ∞ , 61 20 ] . ( − ∞ , 61 20 ] .
Find the domain and range of f ( x ) = 2 ( x − 4 7 ) 2 + 8 11 . f ( x ) = 2 ( x − 4 7 ) 2 + 8 11 .
The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola . We can see the maximum and minimum values in Figure 9.
There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.
A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.
Let’s use a diagram such as Figure 10 to record the given information. It is also helpful to introduce a temporary variable, W , W , to represent the width of the garden and the length of the fence section parallel to the backyard fence.
Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so
A = L W = L ( 80 − 2 L ) A ( L ) = 80 L − 2 L 2 A = L W = L ( 80 − 2 L ) A ( L ) = 80 L − 2 L 2This formula represents the area of the fence in terms of the variable length L . L . The function, written in general form, is
A ( L ) = −2 L 2 + 80 L . A ( L ) = −2 L 2 + 80 L .To find the vertex:
h = − b 2 a k = A ( 20 ) = − 80 2 ( −2 ) and = 80 ( 20 ) − 2 ( 20 ) 2 = 20 = 800 h = − b 2 a k = A ( 20 ) = − 80 2 ( −2 ) and = 80 ( 20 ) − 2 ( 20 ) 2 = 20 = 800
The maximum value of the function is an area of 800 square feet, which occurs when L = 20 L = 20 feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.
This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 11.
Given an application involving revenue, use a quadratic equation to find the maximum.
The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?
Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, p p for price per subscription and Q Q for quantity, giving us the equation Revenue = p Q . Revenue = p Q .
Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently p = 30 p = 30 and Q = 84,000. Q = 84,000. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, p = 32 p = 32 and Q = 79,000. Q = 79,000. From this we can find a linear equation relating the two quantities. The slope will be
m = 79,000 − 84,000 32 − 30 = −5,000 2 = −2,500 m = 79,000 − 84,000 32 − 30 = −5,000 2 = −2,500This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept.
Q = −2500 p + b Substitute in the point Q = 84,000 and p = 30 84,000 = −2500 ( 30 ) + b Solve for b b = 159,000 Q = −2500 p + b Substitute in the point Q = 84,000 and p = 30 84,000 = −2500 ( 30 ) + b Solve for b b = 159,000
This gives us the linear equation Q = −2,500 p + 159,000 Q = −2,500 p + 159,000 relating cost and subscribers. We now return to our revenue equation.
Revenue = p Q Revenue = p ( −2,500 p + 159,000 ) Revenue = −2,500 p 2 + 159,000 p Revenue = p Q Revenue = p ( −2,500 p + 159,000 ) Revenue = −2,500 p 2 + 159,000 p
We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.
h = − 159,000 2 ( −2,500 ) = 31.8 h = − 159,000 2 ( −2,500 ) = 31.8The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.
maximum revenue = −2,500 ( 31.8 ) 2 + 159,000 ( 31.8 ) = 2,528,100 maximum revenue = −2,500 ( 31.8 ) 2 + 159,000 ( 31.8 ) = 2,528,100
This could also be solved by graphing the quadratic as in Figure 12. We can see the maximum revenue on a graph of the quadratic function.
Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the y - y - intercept of a quadratic by evaluating the function at an input of zero, and we find the x - x - intercepts at locations where the output is zero. Notice in Figure 13 that the number of x - x - intercepts can vary depending upon the location of the graph.
Given a quadratic function f ( x ) , f ( x ) , find the y - y - and x-intercepts.
Find the y- and x-intercepts of the quadratic f ( x ) = 3 x 2 + 5 x − 2. f ( x ) = 3 x 2 + 5 x − 2.
We find the y-intercept by evaluating f ( 0 ) . f ( 0 ) .
f ( 0 ) = 3 ( 0 ) 2 + 5 ( 0 ) − 2 = −2 f ( 0 ) = 3 ( 0 ) 2 + 5 ( 0 ) − 2 = −2So the y-intercept is at ( 0 , −2 ) . ( 0 , −2 ) .
For the x-intercepts, we find all solutions of f ( x ) = 0. f ( x ) = 0.
0 = 3 x 2 + 5 x − 2 0 = 3 x 2 + 5 x − 2In this case, the quadratic can be factored easily, providing the simplest method for solution.
0 = ( 3 x − 1 ) ( x + 2 ) 0 = ( 3 x − 1 ) ( x + 2 )So the x-intercepts are at ( 1 3 , 0 ) ( 1 3 , 0 ) and ( − 2 , 0 ) . ( − 2 , 0 ) .
By graphing the function, we can confirm that the graph crosses the y-axis at ( 0 , −2 ) . ( 0 , −2 ) . We can also confirm that the graph crosses the x-axis at ( 1 3 , 0 ) ( 1 3 , 0 ) and ( −2 , 0 ) . ( −2 , 0 ) . See Figure 14
In Example 7, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.
Given a quadratic function, find the x - x - intercepts by rewriting in standard form.
Find the x - x - intercepts of the quadratic function f ( x ) = 2 x 2 + 4 x − 4. f ( x ) = 2 x 2 + 4 x − 4.
We begin by solving for when the output will be zero.
0 = 2 x 2 + 4 x − 4 0 = 2 x 2 + 4 x − 4Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.
f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + kWe know that a = 2. a = 2. Then we solve for h h and k . k .
h = − b 2 a k = f ( −1 ) = − 4 2 ( 2 ) = 2 ( −1 ) 2 + 4 ( −1 ) − 4 = −1 = −6 h = − b 2 a k = f ( −1 ) = − 4 2 ( 2 ) = 2 ( −1 ) 2 + 4 ( −1 ) − 4 = −1 = −6
So now we can rewrite in standard form.
f ( x ) = 2 ( x + 1 ) 2 − 6 f ( x ) = 2 ( x + 1 ) 2 − 6We can now solve for when the output will be zero.
0 = 2 ( x + 1 ) 2 − 6 6 = 2 ( x + 1 ) 2 3 = ( x + 1 ) 2 x + 1 = ± 3 x = − 1 ± 3 0 = 2 ( x + 1 ) 2 − 6 6 = 2 ( x + 1 ) 2 3 = ( x + 1 ) 2 x + 1 = ± 3 x = − 1 ± 3
The graph has x-intercepts at ( −1 − 3 , 0 ) ( −1 − 3 , 0 ) and ( −1 + 3 , 0 ) . ( −1 + 3 , 0 ) .
We can check our work by graphing the given function on a graphing utility and observing the x - x - intercepts. See Figure 15.
We could have achieved the same results using the quadratic formula. Identify a = 2 , b = 4 a = 2 , b = 4 and c = −4. c = −4.
x = − b ± b 2 − 4 a c 2 a = −4 ± 4 2 − 4 ( 2 ) ( −4 ) 2 ( 2 ) = −4 ± 48 4 = −4 ± 3 ( 16 ) 4 = −1 ± 3 x = − b ± b 2 − 4 a c 2 a = −4 ± 4 2 − 4 ( 2 ) ( −4 ) 2 ( 2 ) = −4 ± 48 4 = −4 ± 3 ( 16 ) 4 = −1 ± 3
So the x-intercepts occur at ( − 1 − 3 , 0 ) ( − 1 − 3 , 0 ) and ( − 1 + 3 , 0 ) . ( − 1 + 3 , 0 ) .
In a Try It, we found the standard and general form for the function g ( x ) = 13 + x 2 − 6 x . g ( x ) = 13 + x 2 − 6 x . Now find the y- and x-intercepts (if any).
A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation H ( t ) = − 16 t 2 + 80 t + 40. H ( t ) = − 16 t 2 + 80 t + 40.
k = H ( − b 2 a ) = H ( 2.5 ) = −16 ( 2.5 ) 2 + 80 ( 2.5 ) + 40 = 140 k = H ( − b 2 a ) = H ( 2.5 ) = −16 ( 2.5 ) 2 + 80 ( 2.5 ) + 40 = 140
t = −80 ± 80 2 − 4 ( −16 ) ( 40 ) 2 ( −16 ) = −80 ± 8960 −32 t = −80 ± 80 2 − 4 ( −16 ) ( 40 ) 2 ( −16 ) = −80 ± 8960 −32
Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.
t = − 80 − 8960 − 32 ≈ 5.458 or t = − 80 + 8960 − 32 ≈ − 0.458 t = − 80 − 8960 − 32 ≈ 5.458 or t = − 80 + 8960 − 32 ≈ − 0.458
The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure 16.
a turning point at minus one, three, crosses the x-axis again at the origin, has another turning point at one, minus three, and crosses the x-axis one last time at x = 2, rising from there." width="442" height="372" />
A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation H ( t ) = −16 t 2 + 96 t + 112. H ( t ) = −16 t 2 + 96 t + 112.
Access these online resources for additional instruction and practice with quadratic equations.
Explain the advantage of writing a quadratic function in standard form.
How can the vertex of a parabola be used in solving real-world problems?
Explain why the condition of a ≠ 0 a ≠ 0 is imposed in the definition of the quadratic function.
What is another name for the standard form of a quadratic function?
What two algebraic methods can be used to find the horizontal intercepts of a quadratic function?
For the following exercises, rewrite the quadratic functions in standard form and give the vertex.
f ( x ) = x 2 − 12 x + 32 f ( x ) = x 2 − 12 x + 32
g ( x ) = x 2 + 2 x − 3 g ( x ) = x 2 + 2 x − 3
f ( x ) = x 2 − x f ( x ) = x 2 − x
f ( x ) = x 2 + 5 x − 2 f ( x ) = x 2 + 5 x − 2
h ( x ) = 2 x 2 + 8 x − 10 h ( x ) = 2 x 2 + 8 x − 10
k ( x ) = 3 x 2 − 6 x − 9 k ( x ) = 3 x 2 − 6 x − 9
f ( x ) = 2 x 2 − 6 x f ( x ) = 2 x 2 − 6 x
f ( x ) = 3 x 2 − 5 x − 1 f ( x ) = 3 x 2 − 5 x − 1
For the following exercises, determine whether there is a minimum or maximum value to each quadratic function. Find the value and the axis of symmetry.
y ( x ) = 2 x 2 + 10 x + 12 y ( x ) = 2 x 2 + 10 x + 12
f ( x ) = 2 x 2 − 10 x + 4 f ( x ) = 2 x 2 − 10 x + 4
f ( x ) = − x 2 + 4 x + 3 f ( x ) = − x 2 + 4 x + 3
f ( x ) = 4 x 2 + x − 1 f ( x ) = 4 x 2 + x − 1
h ( t ) = −4 t 2 + 6 t − 1 h ( t ) = −4 t 2 + 6 t − 1
f ( x ) = 1 2 x 2 + 3 x + 1 f ( x ) = 1 2 x 2 + 3 x + 1
f ( x ) = − 1 3 x 2 − 2 x + 3 f ( x ) = − 1 3 x 2 − 2 x + 3
For the following exercises, determine the domain and range of the quadratic function.
f ( x ) = ( x − 3 ) 2 + 2 f ( x ) = ( x − 3 ) 2 + 2
f ( x ) = −2 ( x + 3 ) 2 − 6 f ( x ) = −2 ( x + 3 ) 2 − 6
f ( x ) = x 2 + 6 x + 4 f ( x ) = x 2 + 6 x + 4
f ( x ) = 2 x 2 − 4 x + 2 f ( x ) = 2 x 2 − 4 x + 2
k ( x ) = 3 x 2 − 6 x − 9 k ( x ) = 3 x 2 − 6 x − 9
For the following exercises, use the vertex ( h , k ) ( h , k ) and a point on the graph ( x , y ) ( x , y ) to find the general form of the equation of the quadratic function.
( h , k ) = ( 2 , 0 ) , ( x , y ) = ( 4 , 4 ) ( h , k ) = ( 2 , 0 ) , ( x , y ) = ( 4 , 4 )
f ( x ) = x 2 − 4 x + 4
( h , k ) = ( −2 , −1 ) , ( x , y ) = ( −4 , 3 ) ( h , k ) = ( −2 , −1 ) , ( x , y ) = ( −4 , 3 )
( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 2 , 5 ) ( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 2 , 5 )
( h , k ) = ( 2 , 3 ) , ( x , y ) = ( 5 , 12 ) ( h , k ) = ( 2 , 3 ) , ( x , y ) = ( 5 , 12 )
( h , k ) = ( − 5 , 3 ) , ( x , y ) = ( 2 , 9 ) ( h , k ) = ( − 5 , 3 ) , ( x , y ) = ( 2 , 9 )
f ( x ) = 6 49 x 2 + 60 49 x + 297 49
( h , k ) = ( 3 , 2 ) , ( x , y ) = ( 10 , 1 ) ( h , k ) = ( 3 , 2 ) , ( x , y ) = ( 10 , 1 )
( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 1 , 0 ) ( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 1 , 0 )
( h , k ) = ( 1 , 0 ) , ( x , y ) = ( 0 , 1 ) ( h , k ) = ( 1 , 0 ) , ( x , y ) = ( 0 , 1 )
For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts.
f ( x ) = x 2 − 2 x f ( x ) = x 2 − 2 x
Vertex ( 1 , − 1 ) , Axis of symmetry is x = 1. Intercepts are ( 0 , 0 ) , ( 2 , 0 ) .
f ( x ) = x 2 − 6 x − 1 f ( x ) = x 2 − 6 x − 1
f ( x ) = x 2 − 5 x − 6 f ( x ) = x 2 − 5 x − 6
Vertex ( 5 2 , − 49 4 ) , Axis of symmetry is ( 0 , − 6 ) , ( − 1 , 0 ) , ( 6 , 0 ) .
f ( x ) = x 2 − 7 x + 3 f ( x ) = x 2 − 7 x + 3
f ( x ) = −2 x 2 + 5 x − 8 f ( x ) = −2 x 2 + 5 x − 8
Vertex ( 5 4 , − 39 8 ) , Axis of symmetry is x = 5 4 . Intercepts are ( 0 , − 8 ) .
f ( x ) = 4 x 2 − 12 x − 3 f ( x ) = 4 x 2 − 12 x − 3
For the following exercises, write the equation for the graphed quadratic function.
f ( x ) = x 2 − 4 x + 1
