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)

Answer 1

Answer:

(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)