Question

Write -6x^2+8x-9 in vertex form.

Answer 1

do you know how to complete the square?

Answer 2

the vertex from is \[y=a(x-h)^{2}+k\]

Answer 3

so to bring this we need to complete the square (as phi said)
\[ -6x^2+8x-9\]we set the trinomial equal to y (so that it has 2 "sides" to the equation:\[y=-6x^2+8x-9\]then we move the constant over to the other side:\[y+9=6x^2+8x\]now pull out the GCF of the right side:\[y+9=2(3x^2+4x)\]then find the square of half of the coefficient of x (4x) and add that to the right side:\[y+9=2(3x^2+4x)+4\]then multiply the 4 by the GCF (2):\[y+9=2(3x^2+4x)+8\]now subtract the 9 back over to the other side to isolate y:\[y=2(3x^2+4x)-1\] \[{\large TADA!~:)}\]

Answer 4

no - not standard form

Answer 5

@yummydum Unfortunately your work is not correct.

Answer 6

You could do this
\[ y=-6x^2+8x-9 \]
factor out -6 from the first 2 terms:
\[ y= -6(x^2-\frac{4}{3}x) -9 \]
now add and subtract ((1/2)*4/3)^2 (square of 1/2 the middle term
(this is ok, because we are adding zero)
\[ y= -6(x^2-\frac{4}{3}x +( \frac{2}{3})^2- ( \frac{2}{3})^2) -9 \]
the first 3 terms represent a perfect square
\[ y= -6( (x^2-\frac{4}{3}x +( \frac{2}{3})^2)- ( \frac{4}{3})^2) -9 \]
\[ y= -6( (x-\frac{2}{3})^2- ( \frac{2}{3})^2) -9 \]
now distribute the -6 and simplify to get into standard form
\[ y= -6(x-\frac{2}{3})^2+ 6\cdot ( \frac{4}{9}) -9 \]
\[ y= -6(x-\frac{2}{3})^2- \frac{19}{3}\]