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) (Show Source): Answer by ikleyn(52777) (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 = = -3.
ANSWER. The value of "a" is -3.
Solved.
Answer by MathTherapy(10551) (Show Source):
|
|
|