Exercise 99 Page 200

Practice makes perfect

a From the graph, we can identify the parabola's vertex. This means we can write the function in graphing form.

y= a(x- h)^2+ k [-1em] &Vertex: ( h, k) &Stretch Factor: a Let's identify the coordinates of the function's vertex and x-intercepts.

The vertex is located at (-3,-2) and we have two x-intercepts at (-5,0) and (-1,0). Let's substitute the vertex into the graphing form. y= a(x-( -3))^2+( -2) [-1em] &Vertex: ( -3, -2) &Stretch Factor: a This simplifies to y=a(x+3)^2-2. To find a, we must substitute a second point into the graphing form and then solve for a.

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

SubstituteII

x= -1, y= 0

0=a( -1+3)^2-2

▼

Solve for a

AddTerms

Add terms

0=a(2)^2-2

AddEqn

LHS+2=RHS+2

2=a(2)^2

CalcPow

Calculate power

2=a(4)

RearrangeEqn

Rearrange equation

a(4)=2

DivEqn

.LHS /4.=.RHS /4.

a=2/4

ReduceFrac

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

a=1/2

Now we can complete the graphing form. y= 1/2(x+3)^2-2 [-1em] &Vertex: (-3,-2) &Stretch Factor: 1/2

b From the diagram, we can identify where the line intercepts the x- and y-axis.

To write the equation in slope-intercept form, we need to know the line's slope m, and the y-intercept b. y= mx+ b We know that b= 5. To find the slope, we will substitute our intercepts into the Slope Formula and evaluate.

m = y_2 - y_1/x_2 - x_1

SubstitutePoints

Substitute ( 0,5) & ( -5,0)

m = 0 - 5/- 5 - 0

▼

Evaluate right-hand side

SubTerms

Subtract terms

m = -5/-5

DivNegNeg

- a/- b=a/b

m = 5/5

QuotOne

a/a=1

m = 1

By substituting b= -5 and m= 1 we can write the line's equation. y= 1x+( -5) ⇕ y=x-5

c In previous parts we found the equations of the parabola and of the line. Therefore, we can identify the left-hand side as the equation of the line and the right-hand side as the equation of the parabola.

x+5= 1/2(x+3)^2-2 The solutions to this equation is where the line and parabola intersect. Those are the x-coordinate of the graph's points of intersection

The intersections are (-5,0) and (1,6). Therefore, the solutions to this equation are x=-5 and x=1.

d A solution to a system of equations is a point that satisfies both equations, expressed as (x,y). These points are called points of intersection.

From the diagram, we can identify the solutions to (-5,0) and (1,6).

e The inequality describes where the graph of the line is less than the graph of the parabola. Therefore, we have to identify the x-values where the parabola's graph is above the line's graph. This happens when x is less than -5 or when it's greater than 1.

The solution sets are x<-5 and x>1.

f Solving a quadratic function equal to 0 implies finding the zeros of the parabola. They are located where the graph intercepts the x-axis.

The zeroes are x=-5 and x=-1.

g Here, we are setting the line's right-hand side equal to 4 which means we want to know at which x-value(s) the line has a y-value of 4. We can solve this graphically by drawing the horizontal line y=4 and locating the x-intercept(s) where it intercepts y=x+5.

The solution to the equation is x=-1.

h There are three ways we can transform the function so that it does not intersect the line.

Translate it vertically until the vertex is above the line.
Translate it horizontally until the parabola is to the left of the line.
Reflect it in the horizontal line through the locator point so that the parabola opens downward.

Let's show an example of the first transformation.