Question

(Sketch the graph x^2-7x+12=0​

Answer 1

The graph in the attached figure

Explanation:

we have

`x^(2) -7x+12=0`

This is a vertical parabola open upward (because the leading coefficient is positive)

The vertex is a minimum

step 1

Find the vertex of the quadratic equation

Convert the equation in vertex form

Complete the squares

`(x^(2) -7x+3.5^2)+12-3.5^2=0`

`(x^(2) -7x+12.25)+12-12.25=0`

`(x^(2) -7x+12.25)-0.25=0`

Rewrite as perfect squares

`(x-3.5)^(2)-0.25=0`

The vertex is the point (3.5,0.25)

step 2

Find the x-intercepts

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

we have

`(x-3.5)^(2)-0.25=0`

solve for x

`(x-3.5)^(2)=0.25`

square root both sides

`(x-3.5)=\pm0.50`

`x=3.5\pm0.50`

`x=3.5+0.50=4`

`x=3.5-0.50=3`

therefore

The x-intercepts are the points (3,0) and (4,0)

step 3

Find the y-intercept

The y-intercept is the value of y when the value of x is equal to zero

we have

`y=x^(2) -7x+12`

For x=0

`y=(0)^(2) -7(0)+12`

`y=12`

The y-intercept is the point (0,12)

step 4

Graph the quadratic equation

we have

The vertex (3.5,0.25)

The x-intercepts (3,0) and (4,0)

The y-intercept (0,12)

using a graphing tool

Plot the points and draw the figure

The graph in the attached figure