Exercise 31 Page 206

We have a quadratic function written in standard form, and we want to rewrite it in vertex form. Standard Form y= ax^2+ bx+c Given Equation y=9/4x^2+3x-1 ⇔ y= 9/4x^2+ 3x+(-1) In the given equation, a= 94, b= 3, and c=-1. Let's now recall the vertex form of a quadratic function. Vertex Form: y=a(x-h)^2+k In this equation, a is the leading coefficient of the quadratic function, and the point (h,k) is the vertex of the parabola. By substituting our given values for a and b into the expression - b2a, we can find h.

- b/2a

SubstituteII

a= 9/4, b= 3

- 3/2( 94)

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Simplify

ExpandFrac

a/b=a * 2/b * 2

-6/4* 94

DenomMultFracToNumber

4 * a/4= a

-6/9

ReduceFrac

a/b=.a /3./.b /3.

-2/3

So far, we know that the vertex lies at (- 23,k). To find the y-coordinate k, we will substitute - 23 for x in the given function.

y=9/4x^2+3x-1

Substitute

x= -2/3

y=9/4( -2/3)^2+3( -2/3)-1

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Simplify right-hand side

PowQuot

(a/b)^m=a^m/b^m

y=9/4(4/9)+3(-2/3)-1

MultFracByInverse

a/b* b/a=1

y=1+3(-2/3)-1

DenomMultFracToNumber

3 * a/3= a

y=1-2-1

SubTerms

Subtract terms

y=-2

Therefore, the (h,k) coordinate pair of the vertex is (- 23,-2). Moreover, since we already know that a= 94, we can rewrite the given function in vertex form. y= 9/4 (x-(-2/3))^2+(-2) [1.1em] ⇕ [0.3em] y=9/4(x+2/3)^2-2