How do you find the vertex and the intercepts for $y = -3x^2 + 12x - 8$ ?

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Answer 1 · 2017-04-13T17:11:59

Explained below, which is self explanatory.

$y= -3x^2 +12x -8$

$= -3(x^2 +4x)-8$

$=-3 (x^2 +4x +4 -4) -8$

$= - 3(x^2 +4x +4) +12-8$

$=-3 (x+2)^2 +4$ , is the required vertex form.

This quadratic function represents a vertical parabola opening down.

The vertex is at $(-2,4)$ .

Axis of symmetry is $x= -2$

$y$ - intercept is given by $x=0$ . So it is $-8$

$x$ - intercept is given by $y=0$ .

$x$ -intercepts would be given by the quadratic equation:

$-3x^2 +12x-8=0$ ,

or, $3x^2 -12x +8=0$ ,

$x=(12+- sqrt(144-96))/6" "$ using the quadratic formula.:

$x=(12+-sqrt48)/6$

= $2+-sqrt48/6$

$x = 3.15 and x = 0.85" "$ (to 2 d.p.)

How do you find the vertex and the intercepts for y = -3x^2 + 12x - 8y = -3x^2 + 12x - 8?