Question

values of a, b, and c and the equation of the graph of the parabola
y=ax^2+bx+c
such that is passes through the points
(2,-15)
(-5,-29)
(-3,5)

rewrite it in the form (x-h)^2=4P(y-k)

show all work

Answer 1

Suppose that equation of parabola is
y =ax² + bx + c


Since parabola passes through the point (2,−15) then
−15 = 4a + 2b + c

Since parabola passes through the point (-5,-29), then
−29 = 25a − 5b + c

Since parabola passes through the point (−3,5), then
5 = 9a − 3b + c


Thus, we obtained following system:
4a + 2b + c = −15
25a − 5b + c = −29
9a − 3b + c = 5

Solving it we get that
a = −3, b = −7, c = 11

Thus, equation of parabola is
y = −3x²− 7x + 11

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Rewriting in the form of
(x - h)² = 4p(y - k)

i) -3x² - 7x + 11 = y

ii) -3x² - 7x = y - 11
(-3x² and -7x are isolated)

iii) -3x² - 7x - 147/36 = y - 11 - 147/36
(Adding -147/36 to both sides to get perfect square on LHS)

iv) -3(x² + 7/3x + 49/36) = y - 543/36
(Taking out -3 common from LHS)

v) -3(x + 7/6)² = y - 181/12

vi) (x + 7/6)² = -⅓(y - 181/12)
(Shifting -⅓ to RHS)

vii) (x + 1)² = 4(-1/12)(y - 181/12)
(Rewriting in the form of 4(-1/12) ; This is 4p)


So, after rewriting the equation would be -

(x + 7/6)² = 4(-1/12)(y - 181/12)

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Hope this is what you wanted

- Divyankasc♪