SOLUTION: The graph of a certain quadratic y = ax^2 + bx + c is a parabola with vertex (-4,0) which passes through the point (1,-75). What is the value of a?

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The graph of a certain quadratic y = ax^2 + bx + c is a parabola with vertex (-4,0) which passes through the point (1,-75). What is the value of a?      Log On


   

Question 1189918: The graph of a certain quadratic y = ax^2 + bx + c is a parabola with vertex (-4,0) which passes through the point (1,-75). What is the value of a?
Found 3 solutions by MathLover1, ikleyn, MathTherapy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

The graph of a certain quadratic
y+=+ax%5E2+%2B+bx+%2B+c+
in vertex form"

y+=+%28ax%5E2+%2B+bx%29+%2B+c+
y+=+a%28x%5E2+%2B+%28b%2Fa%29x%2B%28b%2F%282a%29%29%5E2%29+-a%28b%2F%282a%29%29%5E2%2B+c+
y+=+a%28x%5E2+%2B+%28b%2Fa%29x%2B%28b%2F%282a%29%29%5E2%29+-b%5E2%2F4a%2B+c+
y=+a%28x+-+b%2F%282a%29%29%5E2+-b%5E2%2F4a%2B+c+

a parabola with vertex (-4,0) =>h=-4->h=b%2F%282a%29=>b%2F%282a%29=-4->b=-8a
+k=0->0=-b%5E2%2F%284a%29%2B+c+
c=b%5E2%2F4a+->c=%28-8a%29%5E2%2F%284a%29->c=64a%5E2%2F%284a%29->c=16a

y=+a%28x+%2B4%29%5E2+-%28-8a%29%5E2%2F%284a%29%2B+16a
y=+a%28x+%2B4%29%5E2+-16a%2B+16a
y=+a%28x+%2B4%29%5E2++

which passes through the point (1,-75)
-75=+a%281+%2B4%29%5E2

-75=+25a
a=-3

then your equation is:
y=-3%28x+%2B4%29%5E2
y=-3x%5E2+-+24x+-+48




Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
The graph of a certain quadratic y = ax^2 + bx + c is a parabola with vertex (-4,0)
which passes through the point (1,-75). What is the value of a?
~~~~~~~~~~~~~~~~~


            Hello, you do not need make all these calculations and transformations,
            which @MathLover1 makes in her post.

            The solution is in couple of lines below. 


You are given that a quadratic function is a parabola with vertex (-4,0).


It means that the function in vertex form is  

    y = a*(x-(-4))^2 + 0 = a*(x+4)^2.      (1)


The only unknown is the parameter "a". To find it, use the given part, which says
that the parabola passes through the point (1,-75).


So, at x= 1 the value of the function (1) should be -75.  You write this equation, based on (1)

    -75 = a*(1+4))^2,  or  -75 = a*5^2,  which is  25a = -75.


From this equation,  a = -75%2F25 = -3.   


ANSWER.  The value of "a" is  -3.

Solved.



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

The graph of a certain quadratic y = ax^2 + bx + c is a parabola with vertex (-4,0) which passes through the point (1,-75). What is the value of a?
That's ABSOLUTELY unnecessary to complete the square and go through all that "mumbo-jumbo" as the other person did.
This is done as easily as: Vertex form of the equation 
                            of a parabola: matrix%281%2C3%2C+y%2C+%22=%22%2C+a%28x+-+h%29%5E2+%2B+k%29
                                        matrix%281%2C3%2C+-+75%2C+%22=%22%2C+a%281+-+-+4%29%5E2+%2B+0%29 ------ Substituting (1, - 75) for (x, y) and (- 4, 0) for (h, k)
                                         - 75 = 25a
                                       
                                       That's ALL!!