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

Answers ( )

    0
    2022-02-09T04:03:08+00:00

    Answer:

    (3, 0) and (5, 0)

    Step-by-step explanation:

    we have

    [tex]y=x^{2}-8x+15[/tex]

    we know that

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

    so

    For y=0

    [tex]x^{2}-8x+15=0[/tex]

    The formula to solve a quadratic equation of the form

    [tex]ax^{2} +bx+c=0[/tex]

    is equal to

    [tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]

    in this problem we have

    [tex]x^{2}-8x+15=0[/tex]  

    so

    [tex]a=1\\b=-8\\c=15[/tex]

    substitute in the formula

    [tex]x=\frac{-(-8)\pm\sqrt{-8^{2}-4(1)(15)}} {2(1)}[/tex]

    [tex]x=\frac{8\pm\sqrt{4}} {2}[/tex]

    [tex]x=\frac{8\pm2} {2}[/tex]

    [tex]x=\frac{8+2} {2}=5[/tex]

    [tex]x=\frac{8-2} {2}=3[/tex]

    so

    x=3, x=5

    therefore

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

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