Ican't figure this out
this question was on my exam and i got it completely wrong
(i have a graphing calculator capble of matrices so yo don't have to solve the system of equations by hand, use rref (reduced row echelon form))

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
if i were to sub the points in i would ge
(2,-15): -15=4a+2b+c
(-5,-29): -29=25a-5b+c
(-3,5): 5=9a-3b+c

then solve for a, b and c
i don't know how to solve,
(if i don't undestand your answer, i will either report or ask you to explain more)

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 = −2, b = −4, c = 1

Thus, equation of parabola is
y = −2x²− 4x + 1



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

i) -2x² - 4x + 1 = y

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

iii) -3x² - 7x - 49/36 = y - 1 - 49/36
(Adding -49/36 to both sides to get perfect square on LHS)

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

v) -3(x + 7/6)² = y - 445/36 

vi) (x + 7/6)² = -⅓(y - 445/36) 
(Shifting -⅓ to RHS)

vii) (x + 1)² = 4(-1/12)(y - 445/36)
(Rewriting in the form of 4(-1/12) ; This is 4p) 

So, after rewriting the equation would be - 

(x + 7/6)² = 4(-⅛)(y - 445/36)



I hope this is what you wanted.

Regards,
Divyanka♪

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



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)


I have corrected the answer and wrote again since the time to correct the answer from that account was expired.
- Divyanka