# How do you find the x and y intercepts for y=2x^2+8x+10?

May 31, 2016

Set $y = 0$ and then $x = 0$ to find that there are no $x$ intercepts and a single $y$ intercept at $\left(0 , 10\right)$

#### Explanation:

The $x$ and $y$ intercepts of an equation are the points at which $y = 0$ and $x = 0$, respectively. To find them, then, we just plug those values in and solve for the remaining variable.

$x$ intercepts:

Setting $y = 0$, we have

$2 {x}^{2} + 8 x + 10 = 0$

$\implies {x}^{2} + 4 x + 5 = 0$

Noting that the discriminant ${4}^{2} - 4 \left(1\right) \left(5\right) = - 4$ is less than $0$, this has no real solutions. Thus, there are no $x$ intercepts.

$y$ intercepts:

Setting $x = 0$, we have

$y = 2 \cdot {0}^{2} + 8 \cdot 0 + 10$

$\implies y = 10$

Thus there is a single $y$ intercept at $\left(0 , 10\right)$.

If we look at the graph, it appears to agree with our findings:

graph{2x^2+8x+10 [-14.16, 14.32, -0.55, 13.69]}