a,b, and c are constants such that the quadratic ax^2 + bx + c can be expressed in the form 2(x - 5)^2 - 8. When the quadratic 3ax^2 + 3bx + 3c (for the same values of a, b, and c) is expressed in the form n(x - h)^2 + k, what is h?

Guest Jan 5, 2022

#1**+2 **

Answer: 5

Explanation:

(EDIT: The slow way)

I'm going to immediately say that however I solve this is going to be a complete waste of time and that there is almost certainly a smarter way to do this that I can't think of at the moment. But at least I'll get an answer.

Start by expanding \(2(x-5)^2-8\) into the given form:

\(2(x-5)^2-8\)

\(=2(x^2-10x+25)-8\)

\(=2x^2-20x+42\)

It should be clear now that a=2, b=-20, and c=42. So the new expression with the tripled values looks something like this:

\(6x^2-60x+126\)

Now all that needs to be done is converting this into \(n(x-h)^2+k\) form. Already we can tell that n=6.

\(6(x-h)^2+k=6x^2-60x+126\)

\(6x^2-12hx+h^2+k=6x^2-60x+126\)

There is only one term in each side which contains an x^1, namely -12hx and -60x. They must equal each other, and so \(h=5\).

The quicker way:

h didn't change from the two equations, suggesting that there was a quicker way to solve this question.

In this particular question, the x^2 coefficient becomes tripled, the x coefficient becomes tripled, and the integer value becomes tripled.

k is irrelevant to h, so we leave it out of this. Only the \(n(x-h)^2\) part matters to h.

There is one variable present in all three terms of the expanded version of the expression, namely n. So the only number needed to be changed in this part of the form is n. h doesn't change, so if the original value of h was five, the new value is also 5.

I have no idea why I didn't see that before.

WhyamIdoingthis Jan 5, 2022