SOLUTION: The vertex of this parabola is at (2, -4). When the y-value is -3, the x-value is -3. What is the coefficient of the squared term in the parabola's equation?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The vertex of this parabola is at (2, -4). When the y-value is -3, the x-value is -3. What is the coefficient of the squared term in the parabola's equation?      Log On


   

Question 1111615: The vertex of this parabola is at (2, -4). When the y-value is -3, the x-value is -3. What is the coefficient of the squared term in the parabola's equation?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39618) About Me  (Show Source):
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The problem is badly presented. The statement "when the y value is -3, the x value is -3" suggests that y is the independent variable, making the parabola open in the horizontal direction instead of the vertical direction.

So the way I would interpret the problem, the equation of the parabola (vertex form) is
x-2+=+a%28y%2B4%29%5E2

Then we can find the value of a (the coefficient of the squared term, which is what the problem asks for) using the given point (-3,-3) on the parabola.

-3-2+=+a%28-3%2B4%29%5E2
-5+=+a

The coefficient on the squared term is -5; and the equation is x-2+=+-5%28y%2B4%29%5E2.

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Now let's suppose that in fact the intent of the problem was for x to be the independent variable, making the parabola open in the vertical direction.

So now we have the equation
y%2B4+=+a%28x-2%29%5E2

As above, we can find the value of a using the given point (-3,-3):
-3%2B4+=+a%28-3-2%29%5E2
1+=+25a
a+=+1%2F25

If you understand parabolas, there is another way to find the value of a which is, I think, much easier.

In the parabola y=x^2, when you are 1 unit either side of the line of symmetry the function value is 1 more than at the vertex; when you are 2 units either side the function value is 4 more than at the vertex; and so on.

In the parabola y=ax^2, when you are 1 unit either side of the line of symmetry the function value is 1*a=a more than at the vertex; when you are 2 units either side the function value is 4*a=4a more than at the vertex; and so on.

So in this problem, we have the vertex at (2,-4) and a point on the parabola at (-3,-3). The point is 5 units to the left of the vertex; if the coefficient of the squared term were 1, the y value at the given point would be 25 away from the y value at the vertex. But the y value at the given point is only 1 away from the y value at the vertex; that means the coefficient a is 1/25.