Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 4.1
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Exercise 54 Page 184

Practice makes perfect
a To average the x-intercepts we first need to find them. At an x-intercept, the value of y is 0. Therefore, we will set y equal to 0 and solve the equation using the Quadratic Formula.
y=x^2-8x+10
0=x^2-8x+10
x=-( -8)±sqrt(( -8)^2-4( 1)( 10))/2( 1)
Evaluate right-hand side
x=8±sqrt((-8)^2-4(1)(10))/2(1)
x=8±sqrt(8^2-4(1)(10))/2(1)
x=8±sqrt(64-40)/2
x=8±sqrt(24)/2
x=8±sqrt(4(6))/2
x=8±sqrt(4)sqrt(6)/2
x=8±2sqrt(6)/2
x=4±sqrt(6)
lx_1=4-sqrt(6) x_2=4+sqrt(6)

All parabolas are symmetric about their vertex. What this means is if two points have the same y-coordinate, such as the x-intercepts, they are equidistant from the parabola's line of symmetry. Therefore, we can find the line of symmetry by averaging the x-intercepts.

The line of symmetry is x_s=4 which is the x-value of the vertex. We can find the corresponding y-value by substituting 4 for x in the function and evaluating the right-hand side.
y=x^2-8x+10
y= 4^2-8( 4)+10
Evaluate right-hand side
y=16-32+10
y=-6
The vertex is (4,-6).
b By completing the square, we can rewrite the equation in graphing form which allows us to identify the vertex directly from the equation.
Graphing Form:& y=a(x-h)^2+k Vertex:& (h,k)Completing the square requires us to add the square of half the coefficient of x to both sides of the equation.
y=x^2-8x+10
y+ (-8/2)^2=x^2-8x+10+ (-8/2)^2
Solve for y
y+(-4)^2=x^2-8x+10+(-4)^2
y+4^2=x^2-8x+10+4^2
y+4^2=x^2-8x+4^2+10
y+4^2=x^2-2(x)(4)+4^2+10
y+4^2=(x-4)^2+10
y+16=(x-4)^2+10
y=(x-4)^2-6
To identify the vertex, we have to rewrite the function so that it matches the graphing form exactly. Function:& y=(x-4)^2+(-6) Vertex:& (4,-6) The vertex of the quadratic function is (4,-6).
c Let's use the method of averaging the x-intercepts. To find the x-intercepts, we will solve the equation when y=0. To do that, we factor out x and then use the Zero Product Property.
y=x^2-3x
0=x^2-3x
Solve for x
0=x(x-3)
x(x-3)=0
lcx=0 & (I) x-3=0 & (II)
lx_1=0 x_2=3

The x-intercepts have their x-coordinates at x=0 and x=3. By averaging the values, we can find the x-coordinate of the vertex.

The line of symmetry is x_s=1.5 which is the x-value of the vertex. We can find the corresponding y-value by substituting 1.5 for x in our equation.
y=x^2-3x
y= 1.5^2-3( 1.5)
Evaluate right-hand side
y=2.25-4.5
y=-2.25
The vertex of the function is located at (1.5,-2.25).