Hi can someone help me with these grade 12 function question. Please and thank you!

A parabolic arch is used to support a bridge. The arch is 80m wide at the base and the height of the vertex is 20m. How high is the arch at a point 20m in from either end of the base??? lol and answer is 15m

also....
A parabolic dish for picking up sound has a diameter of 1.5 m and is 20cm deep. How high above the center of the dish should the microphone be placed to pick up the sound that is reflected to the focus. Answer is 70.315cm

srry for making it long

2. Originally Posted by swtaznjackie
A parabolic arch is used to support a bridge. The arch is 80m wide at the base and the height of the vertex is 20m. How high is the arch at a point 20m in from either end of the base
You must set it up by drawing a picture. It's a physical problem about a bridge. You should be able to visualize this.

1) Draw a set of coordinate axes, x and y.
2) Put a downward-pointing parabola on the plane defined in 1)

2a) Put the vertex at (0,20)
2b) Label the x-intercepts (40,0) and (-40,0)
2c) Note how the y-axes cuts the parabola in half. The y-axis is the line of symmetry.

3) We need an equation. What do you know about a downward-pointing parabolas with vertex at (0,20)? This should look familiar $4p(y-k) = (x-h)^{2}$. In this case, we know h = 0 and k = 20, so $4p(y-20) = x^{2}$.

4) Solve for the remaining parameter by simple substitution. We have the point (40,0), then $4p(0-20) = 40^{2} \implies p = -20$. I am SO HAPPY this is a negative value. This results in $4(-20)(y-20) = x^{2}$

5) We're ready. What does the problem statement want? 20m from either end of the base. It's symmetric about the y-axis, so it doesn't matter which side we choose. Let's keep it positive. 20m from (40,0) is (20,0), so we need to find the y-coordinate associated with x = 20. Simple substitution and a little algebra and you are done.

Show what you get for that one.

Let's see you tackle the other problem. It's only a test to see if you can find the focus. Use that general equation from part 3.

3. lol so its

4(-20)(y-20)= 400
-80y + 1600 = 400
-80y = -1200
y= 15
omg thank you soo much. Im going to try the other one

4. Originally Posted by swtaznjackie