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# How to solve this parabolic problem?

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Here's the question:

For a), I used the vertex form of the parabola to solve for my equation by taking the min. height of $${4}$$m and the width (10,0) to find my $${a}$$ value.

 $$0 = {a{(10)^2+4}}$$ $$-4 = {100a}$$

making $$a = {-0.04}$$.

plugging this back back into the above equation, i should get the right formula but instead the textbook got $$y = {-0.04x^2 + 5}$$, so i'm confused?

I also need help with b), the answer is $$0 ≤ {-0.04x^{2} + 5}$$ but why include the equal?

Guest Jul 31, 2018
#1
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Here's the question:

a)

$$\begin{array}{|lrcll|} \hline & y &=& a(x-h)^2 + k \quad & | \quad h = 0,\ k = 5 \\ & y &=& a(x-0)^2 + 5 \\ & \mathbf{y} & \mathbf{=}& \mathbf{ax^2 + 5} \\\\ P(5,4): & 4 &=& a5^2 + 5 \\ & 25a + 5 &=& 4 \\ & 25a &=& 4-5 \\ & 25a &=& -1 \quad & | \quad \\ & a &=& -\frac{1}{25} \\ & a &=& -0.04\\\\ \boxed{y=-0.04x^2+5} \\ \hline \end{array}$$

b)

The highway is on level y = 0

heureka  Jul 31, 2018