• June 8th 2007, 06:04 AM
Fibonacci
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!
• June 8th 2007, 06:16 AM
earboth
Quote:

Originally Posted by Fibonacci
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!

Hello,

this is only a guess but...

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

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

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

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

Maybe this helps a little bit.
• June 8th 2007, 06:23 AM
Fibonacci
Quote:

Originally Posted by earboth
Hello,

this is only a guess but...

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

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

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

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

Maybe this helps a little bit.

yah i think thats right but maybe a specific example if u could....
• June 8th 2007, 01:13 PM
earboth
Quote:

Originally Posted by Fibonacci
yah i think thats right but maybe a specific example if u could....

Hello,

$y = -\frac{1}{4} \cdot (x-3)^2 + 4$
$x_{1,2} = 3 \pm \sqrt{\frac{-4}{-\frac{1}{4}}} = 3 \pm 4$