Question

I would appreciate the help please!!

Answer 1

Answer:

the x-intercepts are

$3\ and\ 4$

The vertex is the lowest point on the curve

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

$y=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

if $a>0$ -----> then the parabola open upward (vertex is a minimum)

if $a$ -----> then the parabola open downward (vertex is a maximum)

In this problem we have

$y=x^{2} -7x+12$

$a=1$

so

the parabola open upward (vertex is a minimum)

Find the x-intercepts of the quadratic equation

Equate the equation to zero

$x^{2} -7x+12=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2} -7x=-12$

Complete the square. Remember to balance the equation by adding the same constants to each side

$x^{2} -7x+12.25=-12+12.25$

$x^{2} -7x+12.25=0.25$

Rewrite as perfect squares

$(x-3.5)^{2}=0.25$

square root both sides

$(x-3.5)=(+/-)0.5$

$x=3.5(+/-)0.5$

$x=3.5+0.5=4$

$x=3.5-0.5=3$

therefore

the x-intercepts are

$3\ and\ 4$

Answer 2

X² - 7x +12 = 0
(x-3)(x-4)=0
x= 3 or 4
since the leading coefficient is positive the parabola opens up and the vertex is a minimum