Exercise 52 Page 208

We are given a quadratic function and its vertex, and want to find the values of a and b. Function & Vertex y=ax^2+bx & (-3,2) To do so, we will use the vertex form of a quadratic function. y=a(x- h)^2+ kIn this form, the vertex is ( h, k). Since we know the vertex of our function is ( -3, 2), we can substitute -3 for h and 2 for k. y=a(x-( - 3))^2+ 2 ⇕ y=a(x+3)^2+2 Let's simplify this equation.

y=a(x+3)^2+2

ExpandPosPerfectSquare

(a+b)^2=a^2+2ab+b^2

y=a ( x^2+2(x)(3)+3^2 ) +2

Simplify power and product

y=a ( x^2+6x+9 ) +2

Distr

Distribute a

y=ax^2+6ax+9a+2

Now, we can compare the given function with the one we obtained, term by term.

	y=ax^2+bx+0	y=ax^2+6ax+(9a+2)
Quadratic Coefficient	a	a
Linear Coefficient	b	6a
Independent Term	0	9a+2

We can find the value of a by equating the constant terms.

0= 9a+2

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Simplify

SubEqn

LHS-2=RHS-2

-2=9a

DivEqn

.LHS /9.=.RHS /9.

-2/9 = a

MoveNegNumToFrac

Put minus sign in front of fraction

-2/9 = a

RearrangeEqn

Rearrange equation

a=-2/9

Finally, from the linear coefficient, we know that b= 6a. We will find the value of b by substituting - 29 for a.

b=6a

Substitute

a= -2/9

b=6(- 2/9)

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Simplify right-hand side

MoveLeftFacToNum

a*b/c= a* b/c

b=-12/9

ReduceFrac

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

b=-4/3