Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 months ago

Determine the quadratic function whose vertex is (1, −5) and whose y-intercept is −3. (CRITICAL THINKING)?

Precalc

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  • 6 months ago

    More than one way to skin this cat....

    The y intercept occurs when x = 0. so for this form of the equation:

    y= ax^2 + bx + c

    we have

    -3 = a(0)^2 + b(0) + c 

    or

    -3 = c

    so we know c.

    We also know that is goes throw (1,-5) and so

    -5 = a(1)^2 + b(1) -3

    or  a + b = -2

    The x location of the vertex is the average of the roots.  This makes the

    vertex location  x = -b/(2a)

    So we have  1 = -b/(2a)  which we can switch around to get

    b = -2a

    This can turn a + b = -2 into

    a - 2a = -2 or

    -a = -2

    so a =2

    and since a + b = -2 then 2 + b = -2, and b = -4

    y = 2x^2 - 4x - 3

  • 6 months ago

    Vertex form of a quadratic is:

    y = a(x - h)² + k

    where the vertex is (h, k).  we are told this is (1, -5) so we can substitute those values:

    y = a(x - 1)² - 5

    The y-intercept is the value of y when x = 0.  So we can substitute those two values and can now solve for the remaining unknown:

    -3 = a(0 - 1)² - 5

    -3 = a(-1)² - 5

    -3 = a - 5

    2 = a

    Now our equation is:

    y = 2(x - 1)² - 5

    Finally, expand to get your quadratic in standard form:

    y = 2(x² - 2x + 1) - 5

    y = 2x² - 4x + 2 - 5

    y = 2x² - 4x - 3

  • rotchm
    Lv 7
    6 months ago

    Hint: (h,k) is the vertex, canonical form. Un-anon yourself and I will detail much more.

    Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.  

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