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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:

  1. I divided every term by 5 to get a simplified equivalent function $x^2-4x+3$
  2. 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.

Peter
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    Dividing by 5 guarantees that the $x$-coordinate of the vertex isn't changed, but in general the $y$-coordinate might change. (Draw the picture: multiplying by a constant squeezes/stretches the graph vertically.) Here, the $x$-coordinate is 2, but to get the $y$-coordinate, plug in $x = 2$ into the original function, to get $f(2) = 5(2)^2 - 20(2) + 15 = -5$. – twosigma Feb 26 '21 at 10:00
  • I see, thank you for the clarification :) – Peter Feb 26 '21 at 10:15
  • Anyway, it's a theorem that the vertex of a quadratic function $ax^2+bx+c$ is obtained for $x=-\dfrac b{2a}$. – Bernard Feb 26 '21 at 10:32
  • You made trouble for yourself by not writing the complete curve equation, $ \ y \ = \ 5x^2 -20x + 15 \ \ . $ After dividing through by $ \ 5 \ \ , $ you then have $$ \ \mathbf{\frac{y}{5}} \ = \ x^2 -4x + 3 \ \ . $$ (I've watched many students do this: it is important to be careful about writing complete expressions. I've messed myself up occasionally by not doing that...) –  Jan 06 '22 at 23:43

1 Answers1

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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:

$$ h_g = -1 = \frac{1}{5} * -5 = \frac{1}{5} * h_f . $$

Amaan M
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