Question 638113: Find the x-intercepts of the parabola with vertex (4,75)and y-intercept(o,27). Write your answer in this form: (x1,y1),(x2,y2). If necessary, round to the nearest hundredth.
Found 2 solutions by reviewermath, DrBeeee:
Answer by reviewermath(1029)   (Show Source):
Answer by DrBeeee(684)  (Show Source):
You can put this solution on YOUR website! Given a parabolic function
(1) y = ax^2 + bx +c,
that passes through (0,27) and has a vertex at (4,75).
When x = 0 we have
y = c, therefore
(2) c = 27
The vertex value of x is equal to 4. Using the formula
x = -b/(2a), we get
-b/(2a) = 4 or
(3) b = -8a
Substituting (2) and (3) into (1) yields
(4) y = ax^2 -8ax + 27
Using the vertex point (4,75) we obtain
a4^2 - 8a(4) + 27 = 75 which simplifies to
16a - 32a = 75-27 or
-16a = 48
(5) a = -3
Substitute a into (3) yields
(6) b = 24
The parabola is
(7) y = -3x^2 + 24x + 27
To check (7), put in (4,75)
Is (75 = -3*16 + 96 + 27)?
Is (75 = -48 + 123)?
Is (75 = 75)? Yes
To find the values of x that make y = 0, set (7) to zero
y = -3x^2 +24x +27
y = -3(x^2-8x-9) = 0 or
(x^2 -8x -9) = 0
Factoring y yields
(8) (x-9)(x+1) = 0
Which gives the roots, x = {-1,9}
Check these values by substitution into (7)
Is [0 = -3(-1)^2 +24(-1)+27]?
Is [0 = -3 -24 +27]?
Is [0 = 0]? Yes
Is [0 = -3(9)^2 +24*9+27]?
Is [0 = -243 +216 + 27]?
Is [0 = 0]? Yes
Answer: The x intercepts are {(-1,0),(9,0)}
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