SOLUTION: Hello, how do i fint the vertex and intercepts for this quadratic function: f(x) = x^2

SOLUTION: Hello, how do i fint the vertex and intercepts for this quadratic function: f(x) = x^2 - 6x thank you!

Question 65159This question is from textbook algebra for college students
: Hello,
how do i fint the vertex and intercepts for this quadratic function:
f(x) = x^2 - 6x

thank you! This question is from textbook algebra for college students

Found 3 solutions by josmiceli, funmath, cristiana:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!

The intercepts are (0 , 0) and (6 , 0)
The axis of symmetry will be midway between x = 0 and x = 6
which would be x = 3

So the vertex is at (3 , -9)
Answer by funmath(2933)   (Show Source): You can put this solution on YOUR website!
f(x) = x^2 - 6x
There are a couple of ways to find the vertex, if your teacher is putting the equation in vertex form, let me know and I'll redo this.
When a quadratic equation is in this form: f(x)=ax^2+bx+c, I prefer using this formula to find the x coordinate of the vertex:
a=1, b=-6 and c=0

To find the y-coordinate find f(3)

The vertex (x,y)=(3,-9)
You find the y-intercept, by letting x=0:

The y-intercept is (0,0) The origin will be both an x and a y-intercept.
To find the x-intercept, let f(x)=0 and solve for x.

x=0 and x-6=0
x=0 and x=6
The x-intercepts are (0,0) and (6,0)
Here's what it looks like:

Happy Calculating!!!
Answer by cristiana(10)   (Show Source): You can put this solution on YOUR website!
To easily determine the vertex we have to bring the function to the following form:
and the vertex is: (h,k)
To do that, we basically need to complete the square

Therefore, our new form is:

We can now spot the vertex: (2,9)

To determine the intercepts, we simply have to solve the equation:

Solutions: {0, 6} - these are the x-intercepts

Hope it helped,
Cristiana
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