Quadratics (again)

J4553 · #1 $Old$ November 17th, 2008, 04:23 AM

In the equation y=3t^2 - 7x + 4

I got the x-intercept points as -8 and -8.33

But then got the turning point co-ordinates as 1.17, -0.08
using the x=-b/2a formula then substituting into the quadratic equation.

This doesn't seem like the correct answer, could someone please let me know what I have done wrong?

J4553 · #2 $Old$ November 17th, 2008, 04:26 AM

Wait.. I think I know where I did a wrong calculation... are the x-intercept points 1.33 and 1?

masters · #3 $Old$ November 17th, 2008, 06:20 AM

Quote:

Originally Posted by J4553 $View Post$

In the equation y=3t^2 - 7x + 4

I got the x-intercept points as -8 and -8.33

I must assume you didn't mean for there to be a variable other than x in your equation.

$y=3x^2-7x+4$

$(3x-4)(x-1)=0$

$x=\frac{4}{3} \ \ or \ \ x=1$

Quote:

Originally Posted by J4553 $View Post$

But then got the turning point co-ordinates as 1.17, -0.08
using the x=-b/2a formula then substituting into the quadratic equation.

This doesn't seem like the correct answer, could someone please let me know what I have done wrong?

Unless you are told to do so, I believe I would leave the vertex as fractions rather than rounding them.

V(7/6, -1/12)

J4553 · #4 $Old$ November 17th, 2008, 06:47 AM

Thanks