Exercise 70 Page 552

We need at least three distinct pieces of information from a graph to find the quadratic that fits a parabola. In this case, we are asked to use the following three pieces of information.

1. Axis of Symmetry
2. y-Intercept
3. One x-Intercept

Let's look at the graph and get these pieces of information from it.

Axyis of Symmetry

From the axis of symmetry, we can write use the formula to solve for b in terms of a.

x=- b/2a

Substitute

x= 3

3 = - b/2a

▼

Solve for b

RearrangeEqn

Rearrange equation

- b/2a=3

MultEqn

LHS * 2a=RHS* 2a

- b = 6a

DivEqn

.LHS /(-1).=.RHS /(-1).

b=- 6a

y-Intercept

Now, let's look at the y-intercept on the graph.

From the y-intercept we can conclude that c=16.

One x-Intercept

Let's choose one of the two marked x-intercepts.

We need to choose one zero, it does not matter which one. Let's choose the point (8,0) and substitute those values into our standard form of a quadratic.

y = ax^2 + bx + c

SubstituteII

y= 0, x= 8

0 = a( 8)^2 + b( 8) + c

▼

Solve for a

CalcPow

Calculate power

0 = 64a + 8b + c

Substitute

b= -6a

0= 64a+8(-6a) +c

Multiply

Multiply

0=64a -48a + c

SubTerms

Subtract terms

0 = 16a + c

Substitute

c= 16

0 = 16a + 16

SubEqn

LHS-16a=RHS-16a

-16a=16

DivEqn

.LHS /-16.=.RHS /-16.

a= -1

Now, we can use that and our equation we derived from the axis of symmetry to solve for b. cc b&=&-6 a &⇓& b&=&-6( -1) &⇓& b&=& 6 Since we now have a, b, and c, we can put those values into the standard quadratic form and obtain our function. y= ax^2+ bx+c ⇓ y = - x^2 + 6x + 16