Question

Identify the axis of symmetry, vertex, and y intercept of y=-6x^2+12x-8

Answer 1

Answer:

See below for answers and explanations (with attached graph)

Step-by-step explanation:

It helps to transform the equation into vertex form by completing the square because it tells us a lot about the characteristics of the parabola:

$\displaystyle y=-6x^2+12x-8\\\\y=-6\biggr(x^2-2x+\frac{4}{3}\biggr)\\\\y-6\biggr(-\frac{1}{3}\biggr) =-6\biggr(x^2-2x+\frac{4}{3}-\frac{1}{3}\biggr)\\\\y+2=-6(x^2-2x+1)\\\\y+2=-6(x-1)^2\\\\y=-6(x-1)^2-2$

Since vertex form is $y=a(x-h)^2+k$, we identify the vertex to be $(h,k)\rightarrow(1,-2)$ and the axis of symmetry to be $x=h\rightarrow x=1$. The y-intercept can be found by setting $x=0$ and evaluating:

$y=-6(x-1)^2-2\\\\y=-6(0-1)^2-2\\\\y=-6(-1)^2-2\\\\y=-6(1)-2\\\\y=-6-2\\\\y=-8$

Hence, the y-intercept of the parabola is $y=-8$, or $(0,-8)$ as an ordered pair.