# Find the equation of a Quadratic curve which with a turning point at (-2,3) and which passes through (-1,5)

## Expert Answers Let the quadratic function be f(x) such that:

`f(x)= ax^2 + bx + c`

`==> f(-1) = 5`

`==> f(-1) = a -b + c = 5`.............(1)

Given the turning point is (-1,5) = (Vx,Vy)

Then, we know that:

Vx =-1 = -b/2a ==> b= 2a.........(2)

Vy = 5= -(b^2-4ac)/4a = 5

==> -b^2 +4ac = 20a``

But b= 2a

==> -4a^2 +4ac = 20 a...........(3)

Now, from (1) we know that;

==> a- b +c = 5

==> a - 2a +c = 5

==> -a + c = 5

==> c = a+5 .

Now we will substitute into (3).

==> -4a^2 + 4a(a+5) = 20a

==> -4a^2 +4a^2 +20 = 20a

==> 20 = 20a

==> a = 1

==> b= 2a= 2

==> c = a+5 = 1+5 = 6

==> `f(x)= x^2 + 2x +6 `

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