Find the vertex of the parabola y = 3x2 – 9x + 12, using the formulas for h and k.
a.(-3/2,21/4)
b.(-3/2,-75/4)
c. (-3/2,129/4)
d. (3/2,21/4)
e. none of the above
thank you!!
Hi sky,
What we want to do is put your general equation into vertex form. Let's factor out a 3 from your first two terms like this:
$\displaystyle y=3x^2-9x+12$
$\displaystyle y=3(x^2-3x)+12$
Next, we'll complete the square with what we have inside the parentheses:
$\displaystyle y=3\left(x^2-3x+\frac{9}{4}\right)+12-\frac{27}{4}$
Finally, will arrage the above into vertex form.
$\displaystyle y=3\left(x-\frac{3}{2}\right)^2+\frac{21}{4}$
Vertex form is $\displaystyle y=a(x-h)^2+k$ where (h, k) is the vertex of the parabola.
That would make your vertex $\displaystyle \left(\frac{3}{2}, \frac{21}{4}\right)$