My teacher told me this, but i just want to make sure for my exams on monday, what is the the quadratic formula used to solve x intercepts in an equation when it is in turning point form?

thank you!

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- Jun 8th 2007, 05:04 AMFibonacciQuadratic Formula
My teacher told me this, but i just want to make sure for my exams on monday, what is the the quadratic formula used to solve x intercepts in an equation when it is in turning point form?

thank you! - Jun 8th 2007, 05:16 AMearboth
Hello,

this is only a guess but...

If the vertex has the coordinates V(k, h) the the equation of the parabola is:

$\displaystyle y = a(x-k)^2 + h$ . Equation in turning point form.

To calculate the intercepts with the x-axis you use that y = 0:

$\displaystyle 0 = a(x-k)^2 + h \Longleftrightarrow (x-k)^2 =\frac{-h}{a} \Longleftrightarrow x = k \pm \sqrt{\frac{-h}{a}}$

Maybe this helps a little bit. - Jun 8th 2007, 05:23 AMFibonacci
- Jun 8th 2007, 12:13 PMearboth
Hello,

here comes your example:

$\displaystyle y = -\frac{1}{4} \cdot (x-3)^2 + 4$

The vertex is at V(3, 4) and according to the formula given in my previous post you'll get the zeros Z_1 and Z_2 at:

$\displaystyle x_{1,2} = 3 \pm \sqrt{\frac{-4}{-\frac{1}{4}}} = 3 \pm 4$

I've attached the graph of the parabola so you can check the answer.