We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# How to solve this parabolic problem?

0
254
1

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?

Thank you in advance !

Jul 31, 2018

### 1+0 Answers

#1
+1

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 Jul 31, 2018