The equation of the quadratic function that passes through (-2,1) and (4,-5) with the y-axis as the axis of symmetry is?

2. Originally Posted by Marko_02
The equation of the quadratic function that passes through (-2,1) and (4,-5) with the y-axis as the axis of symmetry is?
$y = ax^2 + bx + c$

y-axis is the axis of symmetry ... $b = 0$

$y = ax^2 + c$

$1 = a(-2)^2 + c$

$-5 = a(4)^2 + c$

solve the system for $a$ and $c$

3. Originally Posted by Marko_02
The equation of the quadratic function that passes through (-2,1) and (4,-5) with the y-axis as the axis of symmetry is?
Let the quadratic be of the form $y = a(x - h)^2 + k$, you can expand it later if you want. now our objective is to find $a,~h, \text{ and }k$.

because we have the points (-2,1) and (4,-5), these are the x-values and corresponding y-values for two points on our parabola.

so we have

$1 = a(-2 - h)^2 + k$ ..............(1) and,
$-5 = a(4 - h)^2 + k$ ..............(2)

now, since the line x = 0 is the axis of symmetry, it means the vertex is on this line. so the x-coordinate for the vertex is 0. since the vertex is (h,k), this means

$h = 0$ ..................(3)

thus you have 3 equations, 3 unknowns, you can solve for the other 2 unknowns to get your quadratic.

it may do you good to see the Conic sections link in my signature

4. Lol thanks for the help but I am new to the course I am still confused by your solutions... I'd be thankful if you can expand further

5. Originally Posted by Marko_02
Lol thanks for the help but I am new to the course I am still confused by your solutions... I'd be thankful if you can expand further
Where did the explanations stop making sense?

You've studied vertices and axes of symmetry in your class, so you know that, if the axis of symmetry is the line x = 0 (that is, the y-axis), then the vertex is of the form (h, k) = (0, k) for some value of "k".

Have you not studied systems of linear equations, so that's where you're bogging down?

Thank you!