Exercise 133 Page 252

a Examining the graph, we see that it exhibits the characteristic of a decreasing exponential function. We see that it has a y-intercept of a= 3 and a horizontal asymptote of k= 0. So far we can write the function as follows.

y= 3b^x+ 0

Let's show this in the graph.

We do not have a second point on the graph. This means we have no way of determining the multiplier. What we do know is that the function is decreasing. This puts the value of b somewhere between 0 and 1. We will arbitrarily choose b= 12. y=3(1/2)^x

b Notice that we have been given two points on the straight line. This is enough to determine its slope m and y-intercept b. Let's consider the slope-intercept form of a straight line.

y= mx+ b We find the slope with the Slope Formula.

m = y_2-y_1/x_2-x_1

SubstitutePoints

Substitute ( 3,- 3) & ( - 3, - 7)

m=-3-( - 7)/3-( - 3)

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Evaluate right-hand side

SubNeg

a-(- b)=a+b

m=-3+7/3+3

AddSubTerms

Add and subtract terms

m=4/6

ReduceFrac

a/b=.a /2./.b /2.

m=2/3

When we know the slope, we can find the y-intercept by substituting one of the points into the slope-intercept form.

y=2/3x+b

SubstituteII

x= 3, y= - 3

- 3=2/3( 3)+b

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Solve for b

FracMultDenomToNumber

a/3* 3 = a

- 3=2+b

SubEqn

LHS-2=RHS-2

- 5=b

RearrangeEqn

Rearrange equation

b=- 5

Now we can finalize the equation of the line. y=2/3x-5

c Examining the equation, we can identify its x-intercepts as x=-3 and x=2 and its y-intercept as y=-6. When we know the x-intercepts of a quadratic function, we can write it in factored form.

FACTORED FORM y= a(x- s)(x- t) [0.1em] [-1.1em] &Vertical Stretch: a &First x-intercept: s &Second x-intercept: t By substituting the x-intercepts into the factored form, we can start writing the function. y= a(x-( -3))(x- 2) [0.1em] [-1.1em] &Vertical Stretch: a &First x-intercept: -3 &Second x-intercept: 2 This simplifies to y=a(x+3)(x-2). We can substitute the y-intercept to find c. Let's do that.

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

SubstituteII

x= 0, y= - 6

- 6=a( 0+3)( 0-2)

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Solve for a

AddSubTerms

Add and subtract terms

- 6=a(3)(-2)

Multiply

- 6=-6a

DivEqn

.LHS /(-6).=.RHS /(-6).

1=a

RearrangeEqn

Rearrange equation

a=1

Now we can finalize the equation of the parabola. y= 1(x-( -3))(x- 2) [0.1em] [-1.1em] &Vertical Stretch: 1 &First x-intercept: -3 &Second x-intercept: 2 This simplifies to y=(x+3)(x-2).

d Let's first have a look at the general equation of a circle.

CIRCLE (x- h)^2+(y- k)^2= r^2 [-1em] & Center: ( h, k) & Radius: r

Examining the circle, we can identify its center and radius.