Question

I need help with my homework. I have the answer but I need to know if my answer is correct please help thank you in advance

Answer 1

EXPLANATION

Since we have the parabola:

$f(x)=6x^2-54$

We first need to obtain the roots in order to compute the x-intercepts:

$\mathrm{Add\: }54\mathrm{\: to\: both\: sides}$$6x^2-54+54=0+54$

Simplify:

$6x^2=54$$\mathrm{Divide\: both\: sides\: by\: }6$$\frac{6x^2}{6}=\frac{54}{6}$

Simplify:

$x^2=9$

For x^2 = f(a) the solutions are x= √f(a) :

$x=\sqrt{9},\: x=-\sqrt{9}$

Computing the square root we get:

$\mathrm{X\: Intercepts}\colon\: \mleft(3,\: 0\mright),\: \mleft(-3,\: 0\mright)$

b) Computing the vertex:

$\mathrm{The\: vertex\: of\: an\: up-down\: facing\: parabola\: of\: the\: form}\: y=ax^2+bx+c\: \mathrm{is}\: x_v=-\frac{b}{2a}$$\mathrm{The\: parabola\: params\: are\colon}$$a=6,\: b=0,\: c=-54$$x_v=-\frac{b}{2a}$$x_v=-\frac{0}{2\cdot\:6}$$\mathrm{Simplify}\colon$$x_v=0$

Plugging in x_v to find the y_v value:

$y_v=6\cdot\: 0^2-54$

Simplify:

$y_v=-54$$\mathrm{Therefore\: the\: parabola\: vertex\: is}\colon$$\mleft(0,\: -54\mright)$