# How do you find the x and y intercepts for y=2x^2-4x+3?

Nov 13, 2015

y-intercept $= 3$
There are no x-intercepts

#### Explanation:

Given $y = 2 {x}^{2} - 4 x + 3$

The y-intercept is the value of $y$ when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} y = 2 {\left(0\right)}^{2} - 4 \left(0\right) + 3 = 3$

For a quadratic in the general form:
$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$
the determinant $\Delta = {b}^{2} - 4 a c$ indicates the number of zeros.

$\Delta \left\{\begin{matrix}< 0 & \Rightarrow & \text{no solutions" \\ =0 & rArr & "one solution" \\ >0 & rArr & "two solutions}\end{matrix}\right.$

In this case
$\textcolor{w h i t e}{\text{XXX}} \Delta = {\left(- 4\right)}^{2} - 4 \left(2\right) \left(3\right) < 0$
so there are no solutions (i.e. no values for which the expression is equal to zero).

This can also be seen from a graph of this equation:
graph{2x^2-4x+3 [-6.66, 13.34, -0.64, 9.36]}

Nov 13, 2015

$\left(0 , 3\right)$

#### Explanation:

$x = 0 \setminus R i g h t a r r o w y = 0 - 0 + 3$

y = 0 \Rightarrow x = frac{-b ± sqrt {b^2 - 4ac}}{2a}

$a = 2 , b = - 4 , c = 3$

But $\Delta$ < 0, then there is no real root $\left({x}_{0} , 0\right)$.