The quadratic function f(x) = ax² + bx + c has tge minimum point (-2,-9) ans f(-1) = -7 Find (a) values of a, b and c i'll mark u brainliest pls, help​

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"katie deleted your answer to the question Hello,a=2, b=8, c=-1Indeed,y=f(x)=ax...

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Answer is a=2, b=8 and c=-1

Indeed:

y=f(x)=ax²+bx+c

Since (-2,-9) is the vertex,

y=k*(x+2)²-9

Or f(-1)=-7 ==> -7=k*(-1+2)²-9 ==> k=2

f(x)=2(x+2)²-9=2*(x²+4x+4)-9=2x²+8x+8-9

f(x)=2x²+8x-1

A proof: the picture is following.

sec ф = 5/3 (quadrant 4)

let x = base

y = perp

z = hypotenuse

sec ф = hypotenuse/base = 5/3

⇒ z = 5 and x = 3

finding y:

z² = x² + y²

5² = 3² + y²

y² = 25-9

y² = 16

y = 4

now finding all other geometric functions of theta:

sin ф = y/z = 4/5

sin ф = -4/5 (sin is negative in the 4th quadrant)

cos ф = x/z = 3/5

cos ф = 3/5 (cos is positive in 4th quadrant)

tan ф = y/x = 4/3

tan ф = -4/3 (tan is negative in 4th quadrant)

cosec ф = z/y = 5/4

cosec ф = -5/4 (cosec is negative in the 4th quadrant)

cot ф = x/y = 3/4

tan ф = -3/4 (tan is negative in 4th quadrant)

idk

step-by-step explanation:

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