Graphing parabolas (making equation )

Question

A parabola has a vertex which is $(1,5)$ and has $x-$intercepts $(a, 0)$ and $(b, 0)$ where $b>0,\ a<0$, and $b = 3|a|$ . solve for $a·b$.

I tried making intercept form but it didnt work. Please help me.

Answer 1

Parabolas are symmetric about their vertex, so $(a,0)$ has to be as far from the vertex as $(b,0)$.

Thus $1-a = b-1$, i.e., $1-a = 3|a|-1$

$\Leftrightarrow |a| = \frac{2-a}{3}$

Think about the graph to solve this:

We know there will be two solutions since the RHS has a positive intercept and negative slope. They already told us $a<0$, so we can substitute $|a|=-a$ and solve:

$$-a = \frac23 - \frac a3$$

So $a = -1$, $b = 3$, and $ab = -3$.

Answer 2

Shortest: Solve

$$ \frac{a+b}{2}=1, \;\;\;\;\; b=3|a| $$

Slightly longer:

$$ y=c(x-1)^2+5 $$

$x=a$ then $0=c(a-1)^2+5$, $x=b$ then $0=c(b-1)^2+5$. So

$$ (b-1)^2=(a-1)^2, \;\;\;\; b=3|a| $$

So you get

$$ b^2-2b=a^2-2a $$

but $b^2=9a^2$ and $2b=6|a|$ so you end up with a quadratic equation in $a$, which you can solve uniquely since $a<0$