Quadratic relations

Question

The x intercepts of a quadratic relation are 0 and 5. The second differences are positive. Explain whether the optimum value is a maximum or a minimum Thank for your help because i don't know how the differences are related to the optimum value. Thank you!

Answer 1

We know that the sum of the roots $-\frac{b}{a}$ is $5$ (1) , and the product $\frac{c}{a}$ (2) is $0$.

In (2) when $a = 0$ the expression is undefined, so that leaves $c = 0$. For (1), there are infinitely many cases where $-b = 5a$. We can consider the cases where $a = 5, b = -1$, and $a = 1, b = -5$ to start with. Therefore, the quadratic equations we have are: $$5x^2-1, -5x^2+1$$

For $5x^2-1$, the first few terms are $4, 19, 44, 79$ and $124$. We can represent the first differences using algebra, by substituting $x$ with $x+1$ for the larger term: $(5(x+1)^2 - 1) - (5x^2 - 1)$, which is:

$$(5x^2+10x+4) - (5x^2-1)$$ $$=10x-5$$

Now we can do the exact same thing, but for the second differences:

$$(10(x+1) - 5) - (10x - 5)$$ $$=(10x+5) - (10x - 5)$$ $$=10$$

and we find out the second differences are all positive. Now, let's look at the first few terms of $-5x^2+1$, which are $-4, -19, -44, 79$ and $-124$. However, since these terms are just the other quadratic's terms multiplied by $-1$, we can conclude that their second differences are all negative.

Since the $x^2$-coefficient of the valid parabola $x^2 - 5x$ is positive, what can you conclude about its vertex?