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Destinyyyy

  • one year ago

Can someone explain how to solve this? Thank you. Find the coordinates of the vertex and the intercepts of the following quadratic function. When necessary, approximate the x-intercepts to the nearest tenth.

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  1. Destinyyyy
    • one year ago
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    f(x)+x^2+6x+5

  2. Destinyyyy
    • one year ago
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    a= 1 b= 6 c= 5 x= -(6)/2(1) =-6/2 =-3

  3. Destinyyyy
    • one year ago
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    @pooja195

  4. princeharryyy
    • one year ago
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    that's the x coordinate of the vertex.

  5. princeharryyy
    • one year ago
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    to find the y coordinate you need to put x =-3 in the equation and solve for the value of y.

  6. Destinyyyy
    • one year ago
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    f(-3)=(-3)^2 -6(-3)+5 = 9+18+5 32

  7. princeharryyy
    • one year ago
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    where y is nothing but f(x) itself. and find the ordered pair as V(x,f(x))

  8. princeharryyy
    • one year ago
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    f(x)=x^2+6x+5 f(-3) = (-3)^2 +6*(-3) +5 = 9-18+5 =4

  9. Destinyyyy
    • one year ago
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    vertex= (-3,-4) ?

  10. princeharryyy
    • one year ago
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    yeah that will be -4 not 4 great, yeah! that's your vertex.

  11. Destinyyyy
    • one year ago
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    Okay... For y-intercept f(0)=(0)^2+6(0)+5 5 (0,5) ?

  12. princeharryyy
    • one year ago
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    yes!

  13. Destinyyyy
    • one year ago
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    Okay

  14. princeharryyy
    • one year ago
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    for x intercepts you need to put x^2+6x+5 = 0 and find all possible x's.

  15. Destinyyyy
    • one year ago
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    Okay so I have to factor x^2+6x+5=0

  16. princeharryyy
    • one year ago
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    yes! (x+1)(x+5) =0; x=-1,-5 and f(x)'s at these x's are 0.

  17. Destinyyyy
    • one year ago
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    Okay.. Is that everything?

  18. princeharryyy
    • one year ago
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    For this question.

  19. Destinyyyy
    • one year ago
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    Okay... Thank you.

  20. princeharryyy
    • one year ago
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    :) wlcm

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