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# Quadratic Functions - PowerPoint PPT Presentation

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The graph of any quadratic function is called a parabola. Parabolas are shaped like cups, as shown in the graph below. If the coefficient of x2 is positive, the parabola opens upward; otherwise, the parabola opens downward. The vertex (or turning point) is the minimum or maximum point.

The Standard Form of a Quadratic Function

f (x)=a(x - h)2+ k, a  0

is in standard form. The graph of f isa parabola whose vertex is the point (h, k). The parabola is symmetric to the line x = h. If a >0, the parabola opens upward; if a< 0, the parabola opens downward.

Graphing Parabolas With Equations in Standard Form

To graph f (x)=a(x - h)2+ k:

• Determine whether the parabola opens upward or downward. If a >0, it opens upward. If a < 0, it opens downward.
• Determine the vertex of the parabola. The vertex is (h, k).
• Find any x-intercepts by replacing f (x) with 0. Solve the resulting quadratic equation for x.
• Find the y-intercept by replacing x with zero.
• Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup.

SolutionWe can graph this function by following the steps in the preceding box. We begin by identifying values for a, h, and k.

Standard formf (x)=a(x - h)2+ k

a = -2 h=3 k=8

Given equationf (x)=-2(x- 3)2+ 8

Text Example

Graph the quadratic function f (x)=-2(x- 3)2+ 8.

Step 1 Determine how the parabola opens. Note that a, the coefficient of x2, is -2. Thus, a < 0; this negative value tells us that the parabola opens downward.

2(x- 3)2= 8

Solve for x. Add 2(x- 3)2 to both sides of the equation.

Text Example cont.

Step 2 Find the vertex. The vertex of the parabola is at (h, k). Because h =3and k =8, the parabola has its vertex at (3, 8).

Step 3 Find the x-intercepts.Replace f(x) with 0 in f(x)=-2(x- 3)2+ 8.

0 =-2(x- 3)2+ 8Find x-intercepts, setting f(x) equal to zero.

(x- 3)2= 4 Divide both sides by 2.

(x- 3) =±2 Apply the square root method.

x- 3= -2 or x- 3= 2 Express as two separate equations.

x= 1 or x= 5Add 3 to both sides in each equation.

The x-interceptsare 1 and 5. The parabola passes through (1, 0) and (5, 0).

Step 4 Find the y-intercept. Replace xwith 0 in f(x)=-2(x- 3)2+ 8.

f(0) =-2(0 - 3)2+ 8=-2(-3)2+ 8 = -2(9) + 8=-10

The y-intercept is –10.The parabola passes through (0, -10).

Text Example cont.

Step 5 Graph the parabola.With a vertex at (3, 8), x-intercepts at 1 and 5, and a y-intercept at –10, the axis of symmetry is the vertical line whose equation is x =3.

The Vertex of a Parabola Whose Equation Isf (x)=ax2+ bx + c

Consider the parabola defined by the quadratic function f (x)=ax 2+ bx + c. The parabola's vertex is at

Example

Graph the quadratic function f (x)=-x2+6x -.

Solution:

Step 1 Determine how the parabola opens. Note that a, the coefficient of x2, is -1. Thus, a < 0; this negative value tells us that the parabola opens downward.

Step 2 Find the vertex. We know the x-coordinate of the vertex is –b/2a.

We identify a, b, and c to substitute the values into the equation for the x-coordinate:

x = -b/(2a) = -6/2(-1) = 3.

The x-coordinate of the vertex is 3. We substitute 3 for x in the equation of the function to find the y-coordinate:

the parabola has its vertex at (3,7).

Example

Graph the quadratic function f (x)=-x2+6x -.

Step 3 Find the x-intercepts.Replace f(x) with 0 in f(x)=-x2+ 6x - 2. 0 = -x2+ 6x - 2

Example

Graph the quadratic function f (x)=-x2+6x -.

Step 4 Find the y-intercept. Replace xwith 0 in f(x)=-x2+ 6x - 2.

f(0) =-02+ 6 • 0 - 2=-

The y-intercept is –2.The parabola passes through (0, -2).

Step 5 Graph the parabola.