Mathematics: Find the missing coefficients y=ax^2-10x+c vertex : (-5,20)

BStep-by-step explanation:The correct answer is B...

Find the missing coefficients y=ax^2-10x+c vertex

paolaviviana

11.03.2022

Remember, for y=ax^2+bx+c

the x value of the vertex is $\frac{-b}{2a}$

so
given
x value of vertex is -5
and
y=ax^2-10x+c

$\frac{-b}{2a}=\frac{-(-10)}{2a}=\frac{10}{2a}=$
$\frac{5}{a}=-5$
multiply both sides by a
5=-5a
divvide both sides by -5
-1=a
a=-1

y=-1x^2-10x+c

comlete the square
remember if we do
y=a(x-h)^2+k

(h,k) is the vertex
I know, we can subsitute the known values of te vertex
-5 for h and 20 for k then expand

y=-1(x-(-5))^2+20

a=-1
c=-5

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Tyrant4life

19.06.2022

Step-by-step explanation:

The correct answer is B

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