Question

(Sketch the graph x^2-7x+12=0​

Answer 1

Answer:

The graph in the attached figure

Step-by-step 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