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

1 Answer
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=-b/(2a)=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(5)=5^2-10*5+5=25-50+5=-20

V=(5, -20)

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

x^2-10x+5 =0 -> x=(10+-sqrt(100-4*1*5))/(2*1) -> x=(10+-sqrt(80))/(2) -> x=5+-2sqrt(5)

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

f(0)=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]}