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
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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)
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I hope this is what you wanted.
Regards,
Divyanka♪
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Answer:
Step-by-step explanation:
Step 1: Define
Difference Quotient: 
f(x) = -x² - 3x + 1
f(x + h) means that x = (x + h)
f(x) is just the normal function
Step 2: Find difference quotient
- <u>Substitute:</u>
![\frac{[-(x+h)^2-3(x+h)+1]-(-x^2-3x+1)}{h}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B-%28x%2Bh%29%5E2-3%28x%2Bh%29%2B1%5D-%28-x%5E2-3x%2B1%29%7D%7Bh%7D)
- <u>Expand and Distribute:</u>
![\frac{[-(x^2+2hx+h^2)-3x-3h+1]+x^2+3x-1}{h}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B-%28x%5E2%2B2hx%2Bh%5E2%29-3x-3h%2B1%5D%2Bx%5E2%2B3x-1%7D%7Bh%7D)
- <u>Distribute:</u>
- <u>Combine like terms:</u>
- <u>Factor out </u><em><u>h</u></em><u>:</u>
- <u>Simplify:</u>
A real number that cannot be expressed as a ratio of integers.
A parallelogram should have 2 sets of parallel lines. Let's find the slope of line PQ and RS to test.
PQ:
(4-2)/(1-(-3))
2/4
1/2
RS:
(2-0)/(3-1)
2/2
1
Because 1 does not equal 1/2 (the slopes are different) the lines are not parallel. Thus, the figure is not a parallelogram.
