Answer to an algebra question

Question

The parabola $y=x^2-3bx-5$, in the $x-y$ plane, has its vertex at $(3,4)$. What is the value of $b$?

I can't answer it. There is no value for $b$ such that, when multiplied by 3, gives 5 and also makes the equation equal 4 when $x$ is 3 (to get the vertex). I'm pretty stumped all around. The problem is from a private SAT prep company. I would appreciate any help.

Answer 1

Hint.

What does "vertex" mean for a parabola? In general, $$ y=a(x-h)^2+k\tag{1} $$ has the vertex $(h,k)$.

Now write the equation of your parabola in the form of (1) by completing the square.

Alternatively, your parabola can be written as $$ y=a(x-3)^2+4\tag{2} $$ for some real number $a$. Expand (2) and compare it with your equation to find $b$.

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^{Following the hint above, you will find that the number $b$ that satisfies the assumptions does not exist.}

Answer 2

I don't think there is any value of $b$ for which it can be the vertex.

Proof:

$4=9-9b-5$ (Vertex lies on the parabola)

$\implies b=0$

But the slope of tangent at the vertex should be zero, that is $\frac{dy}{dx}=0$ at the vertex.

$\implies 2x-3b=0$ at $x=3$

$\implies b=2$ so there isn't any value of $b$ which satisfies that.

Answer 3

It is easiest to see whether or not there is a solution to this problem using graphing.

https://www.desmos.com/calculator/x7cwulidgt

Hint: manipulate the value of $b$ to convince yourself of whether or not there is some value of $b$ for which the parabola with the general equation $y=x^2+3bx-5$ can have a vertex of $P(3,4)$.