I'm in the process of relearning math and came across a problem when calculating the vertex for $5x^2-20x+15$ function. I took the following steps to find the vertex:
- I divided every term by 5 to get a simplified equivalent function $x^2-4x+3$
- Put it in the square form $(x^2 -4x +4) -4 +3$ => $(x-2)^2 -1$
From the $(x-2)^2 -1$ form I would say the vertex coordinates are [2;-1], but I'm wrong. Having looked at wolframalpha it looks likes dividing every term by 5 is a problem, but I don't understand why. Shouldn't it be the same thing? Sorry for the trivial question.
Let's say you have a quadratic function $f(x) = ax^2 + bx + c$. Then, you create a new quadratic function by multiplying by a factor of $\lambda$: $g(x) = \lambda f(x) = \lambda ax^2 + \lambda bx + \lambda c$.
Putting $f$ into vertex form, we have:
$$ y = ax^2 + bx + c $$ $$ y = a \left( x^2 + \frac{b}{a}x \right) + c $$ $$ y = a \left( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right) - \frac{b^2}{4a} + c $$ $$ y = a \left( x - \left(-\frac{b}{2a}\right) \right)^2 - \frac{b^2}{4a} + c $$ $$ y - \left( c - \frac{b^2}{4a} \right) = a \left( x - \left(-\frac{b}{2a}\right) \right)^2 $$
Comparing this to vertex form, $y-k = a(h-k)^2$, we have the following coordinates for the vertex of $f$:
$$ (h_f,k_f) = \left( -\frac{b}{2a} , c - \frac{b^2}{4a} \right) . $$
Now, let's consider what happens to the vertex if we scale our quadratic $f$ by a factor of $\lambda$:
$$ g(x) = \lambda f(x) = \lambda ax^2 + \lambda bx + \lambda c . $$
Following the same process of completing the square, we have:
$$ y = \lambda ax^2 + \lambda bx + \lambda c $$ $$ y = \lambda a \left( x^2 + \frac{b}{a}x \right) + \lambda c $$ $$ y = \lambda a \left( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right) - \frac{\lambda b^2}{4a} + \lambda c $$ $$ y = \lambda a \left( x - \left(-\frac{b}{2a}\right) \right)^2 - \frac{\lambda b^2}{4a} + \lambda c $$ $$ y - \lambda \left( c - \frac{b^2}{4a} \right) = \lambda a \left( x - \left(-\frac{b}{2a}\right) \right)^2 $$ $$ y - \lambda h_f = \lambda a \left( x - k_f \right)^2 $$
Notice, the vertex of $g$ is at coordinates
$$ (h_g, k_g) = (\lambda h_f, k_f) . $$
So, scaling our function by a factor of $\lambda$ leaves the $x$-coordinate of the vertex unchanged, but it does scale the $y$-coordinate of the vertex by that same factor.
In this case, that scaling factor was $\lambda = \frac{1}{5}$, so the $x$-coordinate was the same for both:
$$ h_f = h_g = 2 . $$
But, the $y$-coordinate of the scaled function was $\frac{1}{5}$ the $y$-coordinate of the original function:
$$ h_g = -1 = \frac{1}{5} * -5 = \frac{1}{5} * h_f . $$