Cant figure out a parabola question? (1 Viewer)

[Jayjays332211]

The parabola has a turning point at (z,-8); it intersects the y-axis a y=10 and one of the x-intercepts is x=5. Find:
a) The value of z
b) The equation of the parabola

 

[zeebobDD]

lol you sure this is general mathematics:S

 

[D94]

We know at x = 0, y = 10, and at y = 0, we know one of the intercepts is x = 5.

If we use the general parabolic equation, y = ax² + bx + c, we resolve the above information to get y = ax² + bx + 10. (eqn 2)

We also know the equation of the turning point. It's it's vertex, so we know x = -b/2a. If we make that substitution at y = -8, we now get the expression 0 = 18 - b²/4a. (eqn 3)

Now, we solve for a, which means a = b²/72, and substitute this back into the equation (eqn 2), and letting it equal 0 in order to find b (use quadratic formula).

We then get two results for b. b = -12/5 and -12. Solving for a in (eqn 3), we now have two results for a, which are 2/25 and 2. We must now substitute them back into the equation (eqn 2) and we get two different equations. Apply the information we know (ie. at y = 0, x = 5), it's clear that one equation does not satisfy this.

The correct equation is y = (2/25) x² - (12/5) x + 10. Since the vertex is -b/2a, we find that z = 15.


There might be a faster way, but this is quite straightforward.