Hi,

A quadratic function has zeros −3 and 5 and contains the point (4, 14).

f(x) =  a(x+3)(x-5) 

 14 = a(7)(-1)

 -2 = a

f(x) =  -2(x+3)(x-5) = -2(x^2-2x-15)

f(x) = -2x^2 + 4x + 30 

Wish You the Best in your Studies.

 

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
 .



A quadratic function has zeros −3 and 5 and contains the point (4, 14). Find the equation of the

quadratic function in standard form

~~~~~~~~~~~~~~~~







From the condition, you can write this quadratic function in the form


    q(x) = a*(x-(-3))*(x+5) = a*(x+3)*(x-5)    (*)


with unknown real coefficient "a".


Notice that in this form, the function q(x) has the zeros  at -3 and 5, as prescribed.


Now, to find the unknown coefficient "a", use the condition, that its plot contains the point (4,14).


It means that the value of q(4), when you substitute x= 4 in it, is 14


    14 = a*(4+3)*(4-5),   or

    14 = -7a,

    a  =  = -2.


Therefore, the function q(x) is  q(x) = -2(x+3)*(x-5)}}}.


You may convert it further to the standard form  q(x) = -2x^2 + 4x + 30.    ANSWER




Solved.











 

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