## What are the coordinates of the x-intercepts of the parabola y = x² – 8x + 15? (3, 0) and (5, 0) (3, 0) and (-5, 0) (-3, 0

Question

What are the coordinates of the x-intercepts of the parabola y = x² – 8x + 15?
(3, 0) and (5, 0)
(3, 0) and (-5, 0)
(-3, 0) and (5, 0)
(-3, 0) and (-5, 0)

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3 months 2022-02-09T04:01:41+00:00 1 Answer 0 views 0

(3, 0) and (5, 0)

Step-by-step explanation:

we have

$$y=x^{2}-8x+15$$

we know that

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

so

For y=0

$$x^{2}-8x+15=0$$

The formula to solve a quadratic equation of the form

$$ax^{2} +bx+c=0$$

is equal to

$$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$$

in this problem we have

$$x^{2}-8x+15=0$$

so

$$a=1\\b=-8\\c=15$$

substitute in the formula

$$x=\frac{-(-8)\pm\sqrt{-8^{2}-4(1)(15)}} {2(1)}$$

$$x=\frac{8\pm\sqrt{4}} {2}$$

$$x=\frac{8\pm2} {2}$$

$$x=\frac{8+2} {2}=5$$

$$x=\frac{8-2} {2}=3$$

so

x=3, x=5

therefore

The x-intercepts are (3,0) and (5,0)