Exercise 173 Page 116

Let's let the jump's length in cm be measured on the horizontal axis and the height, also in cm, on the vertical axis. If the origin describes the start of Gloria's jump, we will have two x-intercepts, one at (0,0) and another at (28,0) since she jumps 28 cm.

Since the jump follows a parabola, the maximum height will be at its vertex which is halfway between the x-intercepts. Therefore, the vertex is (14,20). By adding this point to the coordinate plane, we can sketch the graph.

To find an equation for this parabola, we have a few options to choose from. Since we know the graph's x-intercepts, we can write it in factored form. However, we can also write it in graphing form since we know the parabola's vertex. Factored Form:& y=a(x- s)(x- t) x-intercepts:& ( s,0), ( t,0) [1.5em] Graphing Form:& y=a(x-h)+k Vertex:& (h,k) Let's use the graphing form and substitute its vertex. Function:& y=a(x-14)+20 Vertex:& (14,20) We find a by substituting a point on the graph that is not the vertex into the graphing form and solving the resulting equation for a. Let's use the first x-intercept.

y=a(x-14)^2+20

SubstituteII

x= 0, y= 0

0=a( 0-14)^2+20

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Solve for a

SubTerm

Subtract term

0=a(- 14)^2+20

NegBaseToPosBase

(- a)^2=a^2

0=a(14)^2+20

CalcPow

Calculate power

0=a(196)+20

RearrangeEqn

Rearrange equation

a(196)+20=0

SubEqn

LHS-20=RHS-20

a(196)=- 20

DivEqn

.LHS /196.=.RHS /196.

a=- 20/196

ReduceFrac

a/b=.a /4./.b /4.

a=- 5/49

Now we can write the equation. y=- 5/49(x-14)^2+20