Question

In the following exercises, find the x- andy-intercepts.

$y=x^2-4x+10$

Short answer

There is no x- intercept and y- intercept is the point$(0,10)$.

Step-by-step solution

Given information

The given expression is:$y=x^2-4x+10$

Determine the x- intercept.

For xintercept, set $y=0$ and solve forx.

$$\begin{gathered} 0=x^2-4x+10 \\ {x}^2-4x+10=0 \end{gathered}$$

Determine the value of discriminant to predict the number of solution.

$$\begin{gathered} b^2-4ac={(-4)}^2-(-4)\left(1\right)\left(20\right) \\ =16-40 \\ =-26 \end{gathered}$$

Because the value of discriminant is negative, so the equation has no real solution.

There is nox- intercept.

Determine the y- intercept.

Substitute $x=0$and evaluate $y\left(0\right)$.

$$\begin{gathered} y={\left(0\right)}^2-4\left(0\right)+10 \\ =0-0+10 \\ =10 \end{gathered}$$

y- intercept is the point$(0,10)$