Y Intercept

A

asissa

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Sep 6, 2009
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If you are given the vertex of a quadratic function (4, 2) and an x-intercept of (5,0), how is it possible to get the y-intercept? The answer given is (0,-30). It is probably quite simple and I am just not seeing it. I have calculated the slope as -2. I then assumed (rightly or wrongly!) that the function would be y = (x-4)^2 + 2, or x^2 - 8x + 18. Making x = 0 gives me 18??? Help please.
 
Your basic quadratic function has three parameters. You seem to be assuming one of them is one (1). Don't do that.

Start with y-2 = p(x-4)^2 and solve for 'p' using (5,0).
 
Thank you - as soon as I saw the formula I realized my mistake - I had left out the slope, which I had already calculated! p = -2, vertex (4,2)

y-2=p(x-4)^2
y-2=p(x^2 + 8x + 16)
y-2 = -2(x^2 - 8x + 16)
y-2 = -2x^2 + 16x - 32
y=-2x^2 + 16x - 30, fill in zero for x, the y-intercept is (0, -30).

Thank you!
 
Hello, asissa!

Another approach . . .

If you are given the vertex of a quadratic function (4, 2) and an x-intercept of (5,0),
how is it possible to get the y-intercept? .The answer given is (0,-30).
Click to expand...

Assuming that the parbola opens downward, the graph looks like this:

Code: 
    |
    |         (4,2)
    |           o
    |         o : o
    |        o  :  o
    |           :
  - + - - - o - + - o - -
    |       3   4   5
    |           :
    |      o    :    o
    |           :

\(\displaystyle \text{The vertex is at }(4,2).\)
\(\displaystyle \text{One }x\text{-intercept is }(5,0).\)
\(\displaystyle \text{Due to symmetry, the other }x\text{-intercept must be }(3,0).\)

\(\displaystyle \text{The quadratic has the form: }\:y \:=\:a(x-3)(x-5)\)

\(\displaystyle \text{Using }(4,2)\text{ we have: }\:2 \:=\:a(4-3)(4-5) \quad\Rightarrow\quad a \:=\:-2\)

\(\displaystyle \text{Hence: }\:y \:=\:-2(x-3)(x-5) \quad\Rightarrow\quad y \:=\:-2x^2 + 16x - 30\)

\(\displaystyle \text{Therefore, the }y\:\!\text{-intercept is }-30.\)

 
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