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?

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

3. Originally Posted by J4553
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.

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

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

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

Originally Posted by J4553
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)

4. Thanks