A parabola with equation y=ax^2+bx+c has a vertical line of symmetry at x=2 and goes through the two points (1, 1) and (4, -7). The quadratic ax^2+bx+c has two real roots. The greater root is sqrt(n)+2. What is n?

idontknowhowtodivide May 21, 2022

#1**+2 **

A parabola with equation y=ax^2+bx+c has a vertical line of symmetry at x=2 and goes through the two points (1, 1) and (4, -7). The quadratic ax^2+bx+c has two real roots. The greater root is sqrt(n)+2. What is n?

The axis of symetry is just -b/2a

so -b/2a = 2

-b = 4a

b=-2a

So the equation so far becomes y=ax^2 -2ax +c

Sub in your two points and solve simulataneously and what do you get for a and c

Can you take it from here?

Melody May 21, 2022

#2**+1 **

I'm facing another problem now - I got the quadratic equation -8/9x^2+16/9x+1/9=y, but the axis of symmetry is x=1, not x=2.

idontknowhowtodivide
May 21, 2022