Exercise 13 Page 151

To understand the situation, we can graph the equation that models the Gateway Arch. Let's first find the zeros of the equation. Next, we will find its vertex. The mean of the zeros gives the axis of symmetry. With this, we can find the x-coordinate of the vertex. Vertex: 0+635/2=317.5By substituting x=317.5 in the equation, the y-coordinate of the vertex can be found.

y=-0.0063x^2+4x

Substitute

x= 317.5

y=-0.0063( 317.5)^2+4( 317.5)

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Evaluate right-hand side

CalcPow

Calculate power

y=-0.0063(100 806.25)+4(317.5)

Multiply

Multiply

y=-635.07937...+1270

AddTerms

Add terms

y=634.92062...

RoundDec

Round to 0 decimal place(s)

y≈ 635

The vertex is (317.5,635). Now, we can graph the equation.

We want to find the distances where the arch is more than 200 ft above the ground. Let's also graph the line y=200 on the same coordinate plane.

As we can see, between the points of intersection of the line and the parabola, the arch is 200 ft above the ground. To find these points, we should solve the nonlinear system of equations shown below. y=-0.00063x^2+4x & (I) y=200 & (II) To solve the system, we can substitute y=200 in Equation (I) and solve it for x.

y=-0.00063x^2+4x

Substitute

y= 200

200=-0.00063x^2+4x

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Simplify

AddEqn

LHS+0.00063x^2=RHS+0.00063x^2

0.00063x^2+200=4x

SubEqn

LHS-4x=RHS-4x

0.00063x^2-4x+200=0

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Solve using the Zero Product Property

UseQuadForm

Use the Quadratic Formula: a = 0.0063, b= -4, c= 200

x=-( -4)±sqrt(( -4)^2-4( 0.0063)( 200))/2( 0.0063)

Multiply

Multiply

x=-(-4)±sqrt((-4)^2-5.04)/0.0126

NegNeg

- (- a)=a

x=4±sqrt((-4)^2-5.04)/0.0126

CalcPow

Calculate power

x=4±sqrt(16-5.04)/0.0126

AddTerms

Add terms

x=4±sqrt(11.04)/0.0126

CalcRoot

Calculate root

x=4± 3.32264.../0.0126

We can simplify this result into two separate roots.

x=4± 3.32264.../0.0126
x_1=4- 3.32264.../0.0126	x_2=4+ 3.32264.../0.0126
x_1≈ 54.7	x_2≈ 580.2

Therefore, the arch is more than 200 ft above the ground when x∈ (54.7,580.2).