# How do you graph f(x)= x^2 - 10x + 5?

Aug 2, 2015

Calculating the vertex and axis' cut points of your parabola.

#### Explanation:

Since our function is a parabola, to graph it, we will need to know where its vertex is, thus, we apply the vertex calculating formula, which follows:

$x = - \frac{b}{2 a} = \frac{10}{2} = 5$

a, b and c are defined as the coefficients of the variable in the function; $a = 1 , b = - 10 , c = 5$. So we now know that the vertex of the parabola is in $x = 5$, let's calculate now the image of 5 to know exactly where the vertex is.

$f \left(5\right) = {5}^{2} - 10 \cdot 5 + 5 = 25 - 50 + 5 = - 20$

$V = \left(5 , - 20\right)$

Let's calculate where does the function cut the OX axis by equaling the image to 0;

${x}^{2} - 10 x + 5 = 0 \to x = \frac{10 \pm \sqrt{100 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \to x = \frac{10 \pm \sqrt{80}}{2} \to x = 5 \pm 2 \sqrt{5}$

Knowing now the OX cut points, let's find out the OY axis ones:

$f \left(0\right) = 5$

With this information and the fact that $a > 0$ and hence the parabola is convex, we can finally plot:
graph{x^2-10x+5 [-52, 52, -26, 26.03]}