McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
1. Graphing Quadratic Functions
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Exercise 61 Page 226

Determine the vertex of the functions. For f(x), the given graph should be enough to determine its vertex. For g(x), write it in the general form using the given information.

Madison is correct, see solution.

Practice makes perfect

Let's begin by determining the maximums of f(x) and g(x).We are given the following graph of f(x).

We also know that g(x) is a quadratic function with roots of 4 and 2 and a y-intercept of -8. Looking at the graph of f(x), we can immediately determine its vertex as (1,-2).

Therefore, its maximum is -2. To find the maximum of g(x), we should first write it in the general form. g(x)= ax^2+ bx+ c, where a≠0 A quadratic function with roots of x= x_1 and x= x_2 can be written as shown below. g(x)= a(x- x_1)(x- x_2) ⇓ g(x)= a(x- 4)(x- 2) Next, we will find the quadratic coefficient a. The y-intercept of g(x) is given. Therefore, we can find a by substituting x=0 and g(x)=-8 into the function.
g(x)=a(x-4)(x-2)
-8=a( 0-4)( 0-2)
â–Ľ
Solve for a
-8=a(-4)(-2)
-8=8a
-1=a
a=-1
Now we can write the function in the general form by distributing the terms.
g(x)=-1(x-4)(x-2)
â–Ľ
Simplify
g(x)=(- x+4)(x-2)
g(x)=x(- x+4)-2(- x+4)
g(x)=- x^2+4x-2(- x+4)
g(x)=- x^2+4x+2x-8
g(x)=- x^2+6x-8
Now that we wrote the function in the general form, we can determine its vertex to find it maximum. We will first find the x-coordinate of the vertex by using the formula for the axis of symmetry, x=- b2a, where a= -1 and b= 6.
x=-b/2a
x=-6/2( -1)
x=-6/-2
x=6/2
x=3
Next, we will substitute x=3 in the function and find it y-coordinate.
g(x)=- x^2+6x-8
g( 3)=- ( 3)^2+6( 3)-8
g(x)=-9+6(3)-8
g(x)=-9+18-8
g(x)=1
The vertex of g(x) is (1,3). Therefore, its maximum is 1 and it has a greater maximum. As a result, Madison is correct.