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

(3, 0) and (5, 0)

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√(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)