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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?

 May 21, 2022
 #1
avatar+117494 
+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?

 May 21, 2022
 #2
avatar+70 
+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
edited by idontknowhowtodivide  May 21, 2022
 #3
avatar+70 
+1

Just noticed a small error, b=-4a,not -2a. Thanks!

idontknowhowtodivide  May 21, 2022
 #4
avatar+117494 
+1

You are very welcome :)

Melody  May 22, 2022

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