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

1 Answer
Eric Sandin
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]}

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