# The quadratic passes thru the point (-5,8) and the axis of symmetry is x = 3. How do I determine the equation of the quadratic?

Nov 5, 2015

These conditions are satisfied by any quadratic of the form:

$f \left(x\right) = a {\left(x - 3\right)}^{2} + 8 - 64 a = a {x}^{2} - 6 a x + \left(8 - 55 a\right)$

#### Explanation:

Since the axis of symmetry is $x = 3$, the quadratic can be written in the form:

$f \left(x\right) = a {\left(x - 3\right)}^{2} + b$

Since the quadratic passes through $\left(- 5 , 8\right)$ we have:

$8 = f \left(- 5\right) = a {\left(- 5 - 3\right)}^{2} + b = 64 a + b$

Subtract $64 a$ from both ends to get:

$b = 8 - 64 a$

Then:

$f \left(x\right) = a {\left(x - 3\right)}^{2} + 8 - 64 a$

$= a {x}^{2} - 6 a x + 9 a + 8 - 64 a$

$= a {x}^{2} - 6 a x + \left(8 - 55 a\right)$

Here are some of the quadratics that satisfy the conditions:

graph{(x^2-6x-47-y)(1/4x^2-3/2x+8-55/4-y)(-x^2/10+3x/5+13.5-y)=0 [-32.74, 31.35, -11.24, 20.84]}