# How do you find the intercepts, vertex and graph f(x)=x^2-4x-5?

##### 1 Answer
Aug 14, 2017

The $x$ intercepts are at $x = - 1$ and $x = 5$. The $y$ intercept is at $y = - 5$. The vertex is at $\left(2 , 9\right)$.

#### Explanation:

You can use the quadratic formula, the "zero" function on a graphing calculator, factoring, or completing the square to find the solutions. This function is very straightforward to factor. If you're not comfortable with factoring, you can learn about it and practice on Khan Academy. He calls it "factorization": https://www.khanacademy.org/math/algebra/polynomial-factorization.

So $f \left(x\right) = {x}^{2} - 4 x - 5$ becomes $f \left(x\right) = \left(x - 5\right) \left(x + 1\right)$. If you set this equal to zero to get the solutions, which are the points where $f \left(x\right) = 0$ and the function crosses the $x$ axis, you know that either $\left(x - 5\right) = 0$ or $\left(x + 1\right) = 0$.

If $x - 5 = 0$, you can add $5$ to both sides to get $x = 5$.
If $x + 1 = 0$, you can subtract $1$ from both sides to get $x = - 1$.

Finding the vertex is more complex. I like to use the first few steps of completing the square to change the function from "standard form" into "vertex form".
Dr. Khan covers this too, if you're not comfortable with it yet: https://www.khanacademy.org/math/algebra/quadratics/quad-standard-form-alg1/v/ex3-completing-the-square.

Putting this in vertex form yields an equation of the form $y = {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex. Therefore, once you get to $y = {\left(x - 2\right)}^{2} - 9$, you know the vertex is at $\left(h , k\right) = \left(2 , 9\right)$.