Question

Help with those three please!!

Answer 1

Answer:

Part 1) Option A. The axis of symmetry is x=4

Part 2) Option C. minimum

Part 3) Option A. (4,-4)

Part 4) Option B. (2,0) and (6,0)

Step-by-step explanation:

we have

$f(x)=x^{2}-8x+12$

Part 1) What is the axis of symmetry of the parabola described by the equation above?

we know that

The equation above is a vertical parabola open upward

The axis of symmetry is the x-coordinate of the vertex

Find the vertex of the parabola (convert the equation in vertex form)

$f(x)-12=x^{2}-8x$

$f(x)-12+16=x^{2}-8x+16$

$f(x)+4=x^{2}-8x+16$

$f(x)+4=(x-4)^{2}$

$f(x)=(x-4)^{2}-4$ ----> equation in vertex form

The vertex of the parabola is (4,-4)

so

The axis of symmetry is x=4

Part 2) The vertex of the equation above is also

we know that

The equation above is a vertical parabola open upward

therefore

The vertex is a minimum

Part 3) What is the vertex of the parabola described by the equation above

we know that

The equation of the parabola into vertex form is equal to

$f(x)=(x-4)^{2}-4$

therefore

The vertex of the parabola is (4,-4)

Part 4) What are the  x-intercepts of the parabola described by the equation above

we know that

The x-intercepts are the values of x when the value of y is equal to zero

so

$f(x)+4=(x-4)^{2}$

f)x)=0

so

$4=(x-4)^{2}$

take square root both sides

$(x-4)=(+/-)2$

$x=4(+/-)2$

$x=4(+)2=6$

$x=4(-)2=2$

therefore

The x-intercepts are

(2,0) and (6,0)