SOLUTION: A parabola of the form y=ax^2+bx+c has a maximum value of y=3. The y-coordinate of the parabola at x=5 is (9/4). The y-coordinate of the parabola at x=7 is (-15/4). Determine the x

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A parabola of the form y=ax^2+bx+c has a maximum value of y=3. The y-coordinate of the parabola at x=5 is (9/4). The y-coordinate of the parabola at x=7 is (-15/4). Determine the x      Log On


   

Question 1045127: A parabola of the form y=ax^2+bx+c has a maximum value of y=3. The y-coordinate of the parabola at x=5 is (9/4). The y-coordinate of the parabola at x=7 is (-15/4). Determine the x-intercepts of the parabola. Enter your answer in exact form.
Use this form of the equation of a parabola:

y=a(x-h)^2+k
Then h is the x-coordinate of the maximum and k is the maximum value. You'll then need to sub in the two data points to get two equations, which you can solve for a and h or k. Once you've done this you'll have the equation of your parabola, so you can set y = 0 and solve for the x-intercepts.

You'll need to know that the max of a downward facing parabola (negative a) is at the vertex. The vertex has x-coordinate :

xV=(-b/2a)


Smaller X-intercept:
Larger X-intercept:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Two points on the parabola are ( 5, 9/4 ) and ( 7, -15/4 ).
You do not yet know zeros or the actual vertex. You DO know that a%3C0.

All you know about the vertex is some point (h,3). You can say the standard form equation is y=a%28x-h%29%5E2%2B3.


MAKE TWO SPECIFIC EQUATIONS USING THE "GIVEN" POINTS

Still two variables are unknown, being h and a. The two equations using the given points will allow to solve for h and a.

system%28a%285-h%29%5E2%2B3=9%2F4%2Ca%287-h%29%5E2%2B3=-15%2F4%29

system%28a%285-h%29%5E2=9%2F4-12%2F4%2Ca%287-h%29%5E2=-15%2F4-12%2F4%29

system%28a%285-h%29%5E2=-3%2F4%2Ca%287-h%29%5E2=-27%2F4%29

Find the two expressions for "a".

system%28a=-3%2F%284%285-h%29%5E2%29%2Ca=-27%2F%284%287-h%29%5E2%29%29

4%285-h%29%5E2%2F3=4%287-h%29%5E2%2F27

27%2A4%285-h%29%5E2%2F3=27%2A4%287-h%29%5E2%2F27

36%285-h%29%5E2=4%287-h%29%5E2

9%285-h%29%5E2=%287-h%29%5E2

9%2825-10h%2Bh%5E2%29=49-14h%2Bh%5E2

225-90h%2B9h%5E2=49-14h%2Bh%5E2

225-49-90h%2B14h%2B8h%5E2=0

176-76h%2B8h%5E2=0

highlight_green%284h%5E2-19h%2B44=0%29
D=19%5E2-4%2A4%2A44=19%5E2-16%2A44=361-704----------not working; unknown mistake.


I am leaving this incomplete and possibly incorrect solution posted anyway. Maybe you understand the path taken and may do it properly and avoid whatever mistake I have made.


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SHOULD be, the quadratic equation in just h,
highlight_green%282h%5E2-19h%2B44=0%29
and then discriminant, D=19%5E2-4%2A2%2A44=361-352=9=highlight_green%283%5E2%29, giving solution for h as
h=%2819%2B-+3%29%2F%282%2A2%29
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h=%2819%2B-+3%29%2F4
system%28h=4%2Cor%2Ch=11%2F2%29-----and maybe one of this might not work.

CORRESPONDING "a"?
Using either of the earlier equations solved for a,
a=-3%2F%284%285-h%29%5E2%29
-
a=-3%2F%284%285-4%29%5E2%29
a=-3%2F%284%2A1%29
a=-3%2F4
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a=-3%2F%284%285-11%2F2%29%5E2%29
a=-3%2F%284%2810%2F2-11%2F2%29%5E2%29
a=-3%2F%284%28-1%2F2%29%5E2%29
a=-3%2F%284%281%2F4%29%29
a=-3
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This gives two combinations of solutions of this system for a and h.
Either system%28a=-3%2F4%2Cand%2Ch=4%29
OR system%28a=-3%2Cand%2Ch=11%2F2%29.


WHAT IS THE STANDARD FORM EQUATION USING THESE?

Starting from y=a%28x-h%29%5E2%2B3,

-------------------------------------------------------
system%28y=-%283%2F4%29%28x-4%29%5E2%2B3%2Cor%2Cy=-3%28x-11%2F2%29%5E2%2B3%29
-------------------------------------------------------


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The question asked was, what are the y-intercepts. That not actually answered, but you could now easily just let y=0 and solve for x. You may still need to do both equations separately.