How do you find the x and y intercepts for x^2-6x+1?

May 17, 2016

Intercepts on $x$ axis are $\left(3 + 2 \sqrt{2} , 0\right)$ and $\left(3 - 2 \sqrt{2} , 0\right)$

and intercept on $y$ axis is $\left(0 , 1\right)$.

Explanation:

For finding $x$ and $y$ intercepts of $y = {x}^{2} - 6 x + 1$

Let us first put $y = 0$ (to find intercepts on $x$ axis)

or ${x}^{2} - 6 x + 1 = 0$.

Using quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, we get

$x = \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \cdot 1 \cdot 1}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2 \sqrt{2}$

Hence, intercepts on $x$ axis are $\left(3 + 2 \sqrt{2} , 0\right)$ and $\left(3 - 2 \sqrt{2} , 0\right)$

For intercepts on $y$ axis, put $x = 0$ i.e.

$y = {0}^{2} - 6 \cdot 0 + 1 = 1$

Hence, intercept on $y$ axis is $\left(0 , 1\right)$.

graph{x^2-6x+1 [-8.5, 11.5, -5.64, 4.36]}