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Effectus [21]
3 years ago
10

Write the equation for a parabola whose vertex is (-6,-1) and passes through (-12,17).

Mathematics
1 answer:
Hitman42 [59]3 years ago
7 0

Answer:

y=\frac{1}{2}(x+6)^2-1

Step-by-step explanation:

<u>Equation of the Quadratic Function</u>

The vertex form of the quadratic function has the following equation:

y=a(x-h)^2+k

Where (h, k) is the vertex of the parabola that graphically represents the function, and a is a coefficient different from zero.

The vertex of the required equation is located at (-6,-1)

The parabola passes through (-12,17)

Substituting the coordinates of the vertex, the equation of the function is:

y=a(x+6)^2-1

The value of a will be determined by using the given point:

17=a(-12+6)^2-1

Operating:

17=a(36)-1

36a=18\Rightarrow a=18/36=1/2

Solving:

a=1/2

The equation of the parabola is:

\boxed{y=\frac{1}{2}(x+6)^2-1}

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================================================

Work Shown:

Compute the function value at the endpoints

f(x) = \sqrt{4-x}\\\\f(-5) = \sqrt{4-(-5)} = 3\\\\f(4) = \sqrt{4-4} = 0\\\\

With a = -5 and b = 4, we have

f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(4)-f(-5)}{4-(-5)}\\\\f'(c) = \frac{0-3}{9}\\\\f'(c) = -\frac{1}{3}\\\\

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Use algebra to solve for c

-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\\frac{1}{2\sqrt{4-c}} = \frac{1}{3}\\\\3 = 2\sqrt{4-c}\\\\2\sqrt{4-c} = 3\\\\\sqrt{4-c} = \frac{3}{2}\\\\4-c = \frac{9}{4}\\\\c = 4-\frac{9}{4}\\\\c = \frac{16-9}{4}\\\\c = \frac{7}{4}\\\\

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3 years ago
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