Draw a diagram of the archway modeled by the equation y = x^2 + 5x + 24. Fine and label the y-intercept and the x-intercepts on the sketch. Then find and label the width of the archway at its base and the height of the archway at its highest point, assuming the base of the archway is along the x-axis ​

Draw A Diagram Of The Archway Modeled By The Equation Y = X^2 + 5x + 24. Fine And Label The Y-intercept

Answers

Answer 1

Answer:

1) Please find the attached drawing of the archway created with MS Excel

2) The y-intercept is (0, 24)

The x-intercepts are (-3, 0), and (8, 0)

3) The width of the archway at its base is 11

The height of the archway at its highest point = 30.25

Step-by-step explanation:

1) Please find the attached drawing of the archway created with MS Excel

2) The y-intercept is (0, 24)

The x-intercepts are (-3, 0), and (8, 0)

3) The width of the archway at its base is 11

The height of the archway at its highest point = 30.25

Step-by-step explanation:

The given function representing the archway is y = -x² + 5·x + 24

1) Please find attached the required drawing of the archway created with MS Excel

2) The y-intercept is given by the point where x = 0

Therefore, we have, the y-value at the y-intercept = -0² + 5×0 + 24 = 24

The y-intercept = (0, 24)

The x-intercept is given by the point where y = 0

Therefore, the x-values at the x-intercept are found using the following equation;

0 = -x² + 5·x + 24

x² - 5·x - 24 = 0

By inspection, we have;

x² - 8·x + 3·x - 24 = 0

x·(x - 8) + 3·(x - 8) = 0

∴ (x + 3) × (x - 8) = 0

Either (x + 3) = 0, and x = -3, or (x - 8) = 0, and x = 8

Therefore, the x-intercepts are (-3, 0), and (8, 0)

3) The width of the archway at its base = The distance between the x-values at the two x-intercepts

∴ The width of the archway at its base = 8 - (-3) = 11

The highest point of the arch is given by the vertex of the parabola, y = a·x² + b·x + c, which has the x-value of the vertex = -b/(2·a)

∴ The x-value of the vertex of the given parabola, y = -x² + 5·x + 24, is x = -5/(2×(-1)) = 2.5

Therefore;

The y-value of the vertex, is y = -(2.5)² + 5×2.5 + 24 = 30.25 = The height of the archway at its highest point

∴ The height of the archway at its highest point = 30.25

Draw A Diagram Of The Archway Modeled By The Equation Y = X^2 + 5x + 24. Fine And Label The Y-intercept
Answer 2

The vertex of the parabola is at (-2.5, 17.75). And the y-intercept of the parabola is at (0, 24).

What is the equation of the parabola?

Let the point (h, k) be the vertex of the parabola and a be the leading coefficient.

Then the equation of the parabola will be given as,

y = a(x - h)² + k

The equation of the parabola is given as,

y = x² + 5x + 24

Convert the equation into a vertex form, then we have

y = x² + 5x + 25/4 - 25/4 + 24

y = (x + 5/2)² + 71/4

y = (x + 2.5)² + 17.75

The vertex of the parabola is at (-2.5, 17.75). The y-intercept of the parabola is given as,

y = (0)² + 5(0) + 24

y = 24

The y-intercept of the parabola is at (0, 24).

More about the equation of the parabola link is given below.

https://brainly.com/question/20333425

#SPJ2

Draw A Diagram Of The Archway Modeled By The Equation Y = X^2 + 5x + 24. Fine And Label The Y-intercept

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Looking at the first system of equations,

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If we multiply both sides of the second equation by 2, the coefficient of x is exactly the negative of the coefficient of x in the first equation.

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The main idea behind elimination is combining the given equations in just the right amount so that one of the variables disappears. The "right amount" involves using the LCM of the coefficients of a given variable. In this example, the x-coefficients had LCM(8, 16) = 16, so we only had to scale one of the equations (the one with -8x) to cancel all the x terms.

If we wanted to eliminate y first instead, we first note that LCM(6, 10) = 30. To get 30 as a coefficient on y, in the first equation we would have multiplied by 3:

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And in the second equation, we would have multiplied by -5 (negative so that upon combining the equations, we end up with -30y + 30y = 0):

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16x - 10y = 10

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Step-by-step explanation:

1) I want to talk about reflections first.

Reflections across the x-axis --> [tex]y = ax^2[/tex], a is the coefficient. if a is negative, then the equation should be reflected across the x-axis. This is known as a vertical reflection. Reflections across the y-axis --> [tex]y=a(bx)^2[/tex], b is the coefficient. If b is negative, then reflect the equation over the y-axis. There are cases where the reflection across the y-axis does not change anything. But, let's say its [tex]y=(x-3)^2[/tex]... the reflection across the y-axis is different (that equation is: [tex]y=(-x-3)^2[/tex] )

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Example (check image for visual)

We transform [tex]y = x^2[/tex] to [tex]y = -(-x-3)^2+3[/tex] , you move right 3, then reflect across the x-axis, then reflect across y-axis, then move 3 up.

--------------------------------------------------------------------------------------------------------------

Note: In the image, the red line is the original function, the blue one is the transformed function. See if you can follow along with the verbal instructions I gave above.

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