Introduction We have studied the key features of the graph of a parabola, such as the vertex and x-intercepts. In this lesson, we will review the definitions

IntroductionWe have studied the key features of the graph of a parabola, such as the vertex and x-intercepts. In this lesson, we will review the definitions of key features and learn to identify them from the equation of a quadratic function. Once you identify the key features, you can make a sketch of the graph of a quadratic equation.

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5.6.1: Graphing Quadratic Functions

Key Concepts• A quadratic function in standard form, or general form,

is written as f(x) = ax2 + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term.

• The graph of a quadratic is U-shaped and called a parabola.

• The extremum of a graph is the function value that achieves either a maximum or a minimum.

• The maximum is the largest y-value of a quadratic, and the minimum is the smallest y-value of a quadratic.

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Key Concepts, continued• If a > 0, the parabola is concave up and the quadratic

has a minimum. • If a < 0, the parabola is concave down and the

quadratic has a maximum. • The extreme values of a quadratic occur at the vertex,

the point at which the curve changes direction.

• If the quadratic equation is given in standard form, the

vertex can be found by identifying the x-coordinate of

the vertex using

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Key Concepts, continued• Substitute the value of the x-coordinate into the

quadratic equation to find the y-coordinate of the vertex.

• The vertex of a quadratic function is

• Another way to find the vertex is to complete the square on the standard form of the parabola in order to convert it to vertex form.

• The vertex form of a quadratic function is f(x) = a(x – h)2 + k, where the coordinate pair (h, k) is the location of the vertex.

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Key Concepts, continued• The intercept of a graph is the point at which a line

intercepts the x- or y-axis. • The y-intercept of a function is the point at which the

graph crosses the y-axis. This occurs when x = 0. The y-intercept is written as (0, y).

• The x-intercepts of a function are the points at which the graph crosses the x-axis. This occurs when y = 0. The x-intercept is written as (x, 0).

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Key Concepts, continued• The zeros of a function are the x-values for which the

function value is 0. • The intercept form of the quadratic function, written as

f(x) = a(x – p)(x – q), where p and q are the zeros of the function, can be used to identify the x-intercepts. Set the factored form equal to 0. Then set each factor equal to 0. As long as the coefficients of x are 1, the x-intercepts are located at (r, 0) and (s, 0).

• These values for x are the roots, or the solutions, of the quadratic equation.

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Key Concepts, continued• Parabolas are symmetrical; that is, they have two

identical parts when rotated around a point or reflected over a line.

• This line is the axis of symmetry, the line through the vertex of a parabola about which the parabola is symmetric. The equation of the axis of symmetry is

• Symmetry can be used to find the vertex of a parabola if the vertex is not known.

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Key Concepts, continued• If you know the x-intercepts of the graph or any two

points on the graph with the same y-value, the x-coordinate of the vertex is the point halfway between the values of the x-coordinates.

• For x-intercepts (r, 0) and (s, 0), the x-coordinate of

the vertex is

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Key Concepts, continued

Key Concepts, continued• To graph a function using a graphing calculator, follow

these general steps for your calculator model. On a TI-83/84:

Step 1: Press the [Y=] button.

Step 2: Type the function into Y1, or any available equation. Use the [X, T, θ, n] button for the variable x. Use the [x2] button for a square.

Step 3: Press [WINDOW]. Enter values for Xmin, Xmax, Ymin, and Ymax. The Xscl and Yscl are arbitrary. Leave Xres = 1.

Step 4: Press [GRAPH]. 10


Key Concepts, continuedOn a TI-Nspire:

Step 1: Press the [home] key.

Step 2: Arrow over to the graphing icon and press [enter].

Step 3: Type the function next to f1(x), or any available equation, and press [enter]. Use the [X] button for the letter x. Use the [x2] button for a square.

Step 4: To change the viewing window, press [menu]. Select 4: Window/Zoom and select A: Zoom – Fit.

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Common Errors/Misconceptions• incorrectly replacing x with 0 instead of y when

determining the x-intercept (and vice versa) • using the incorrect sign for the x-coordinate of the

vertex when using vertex form • using the incorrect sign when identifying the intercepts

from the factored form of the quadratic

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Guided Practice

Example 1Given the function f(x) = –2x2 + 16x – 30, identify the key features of the graph: the extremum, vertex, x-intercept(s), and y-intercept. Then sketch the graph.

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Guided Practice: Example 1, continued

1. Determine the extremum of the graph. The extreme value is a minimum when a > 0. It is a maximum when a < 0.

Because a = –2, the graph opens downward and the quadratic has a maximum.

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2. Determine the vertex of the graph. The maximum value occurs at the vertex.

The vertex is of the form

Use the original equation f(x) = –2x2 + 16x – 30 to find the values of a and b in order to find the x-value of the vertex.

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The x-coordinate of the vertex is 4.

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Formula to find thex-coordinate of the vertex of a quadratic

Substitute –2 for a and 16 for b.

x = 4 Simplify.

Guided Practice: Example 1, continuedSubstitute 4 into the original equation to find the y-coordinate.

The y-coordinate of the vertex is 2.

The vertex is located at (4, 2).

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f(x) = –2x2 + 16x – 30 Original equation

f(4) = –2(4)2 + 16(4) – 30 Substitute 4 for x.

f(4) = 2 Simplify.


3. Determine the x-intercept(s) of the graph.Since the vertex is above the x-axis and the graph opens downward, there will be two x-intercepts.

Factor the quadratic and set each factor equal to 0.

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The x-intercepts are (3, 0) and (5, 0). 19



f(x) = –2(x2 – 8x + 15)Factor out the greatest common factor.

f(x) = –2(x – 3)(x – 5) Factor the trinomial.

0 = –2(x – 3)(x – 5)Set the factored form equal to 0 to find the intercepts.

x – 3 = 0 or x – 5 = 0Set each factor equal to 0 and solve for x.

x = 3 or x = 5 Simplify.


4. Determine the y-intercept of the graph.The y-intercept occurs when x = 0.

Substitute 0 for x in the original equation.

The y-intercept is (0, –30).

When the quadratic equation is written in standard form, the y-intercept is c.

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f(0) = –30 Simplify.


5. Graph the function.Use symmetry to identify additional points on the graph.

The axis of symmetry goes through the vertex, so the axis of symmetry is x = 4.

For each point to the left of the axis of symmetry, there is another point the same distance on the right side of the axis and vice versa.

The point (0, –30) is on the graph, and 0 is 4 units to the left of the axis of symmetry.

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Guided Practice: Example 1, continuedThe point that is 4 units to the right of the axis is 8, so the point (8, –30) is also on the graph.

Determine two additional points on the graph.

Choose an x-value to the left or right of the vertex and find the corresponding y-value.

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f(1) = –16 Simplify.

Guided Practice: Example 1, continuedAn additional point is (1, –16).

(1, –16) is 3 units to the left of the axis of symmetry.

The point that is 3 units to the right of the axis is 7, so the point (7, –16) is also on the graph.

Plot the points and join them with a smooth curve.

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Guided Practice

Example 3Given the function f(x) = –2(x + 1)(x + 5), identify the key features of its graph: the extremum, vertex, x-intercept(s), and y-intercept. Then sketch the graph.

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1. Determine the extremum of the graph. The extreme value is either a minimum, when a > 0, or a maximum, when a < 0.

Because a = –2, the graph opens down and the quadratic has a maximum.

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2. Determine the x-intercept(s) of the graph. The equation is in factored form.

The zeros occur when y = 0.

Find the zeros by setting the equation equal to 0.

Then set each factor equal to 0.

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The x-intercepts are (–1, 0) and (–5, 0).

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f(x) = –2(x + 1)(x + 5) Original equation

0 = –2(x + 1)(x + 5) Set the equation equal to 0.

x + 1 = 0 or x + 5 = 0

x = –1 or x = –5


3. Determine the vertex of the graph.Use the axis of symmetry to identify the vertex.

The axis of symmetry is halfway between the x-intercepts.

Find the midpoint between the x-intercepts.

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The axis of symmetry is x = –3, so the x-coordinate of the vertex is –3.

To find the y-coordinate, substitute –3 into the original equation.

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Midpoint formula

Substitute –1 for x1 and –5 for x2.

Simplify.


The vertex is (–3, 8).

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f(–3) = –2[(–3) + 1][(–3) + 5] Substitute –3 for x.

f(–3) = –2(–2)(2) Simplify.

f(–3) = 8


4. Determine the y-intercept of the graph.The y-intercept occurs when x = 0.

Substitute 0 for x in the original equation.

The y-intercept is (0, –10).

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f(0) = –2[(0) + 1][(0) + 5] Substitute 0 for x.

f(0) = –2(1)(5) Simplify.

f(0) = –10


5. Graph the function.Use symmetry to identify additional points on the graph.

Because 0 is 3 units to the right of the axis of symmetry, the point 3 units to the left of the axis will have the same value, so (–6, –10) is also on the graph.

Determine two additional points on the graph.

Choose an x-value to the left or right of the vertex and find the corresponding y-value.

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An additional point is (–2, 6).

(–2, 6) is 1 unit to right of the axis of symmetry.

The point that is 1 unit to the left of the axis is –4, so the point (–4, 6) is also on the graph.

Plot the points and join them with a smooth curve. 35



f(–2) = –2[(–2) + 1][(–2) + 5]Substitute –2 for x.

f(–2) = 6 Simplify.


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Documents

Introduction We have studied the key features of the graph of a parabola, such as the vertex and x-intercepts. In this lesson, we will review the definitions