Question

I would appreciate the help please!!

The graph of y = x2 –7x +12 is a parabola with x-intercepts at: {-3 and -4, or 3 and 4} and a vertex that is the: {lowest or highest} point on the curve.

Thanks in advance!

Answer 1

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

Answer 2

the x-intercepts are

`3\ and\ 4`

The vertex is the lowest point on the curve

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<0` -----> 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`