Answer
The $y$-intercept is $1$.
The $x$-intercepts are $\dfrac{2-\sqrt 6}{2}$ and $\dfrac{2+\sqrt 6}{2}$.
Work Step by Step
Determine the $y$-intercept by setting $x=0$ then solving the equation for $y$:
$y=f(0)=-2(0^2)+4(0)+1=1$
Thus, the $y$-intercept is $1$.
Determine the $x$-intercepts by sestting $y$ to $0$ then solving the equation using the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.
$-2x^2+4x+1=0$ (so $a=-2, b=4,$ and $c=1$)
$x=\dfrac{-4\pm\sqrt{4^2-4(-2)(1)}}{2(-2)}=\dfrac{-4\pm\sqrt{24}}{-4}\\
x=\dfrac{-4\pm\sqrt{4\cdot6}}{-4}=\dfrac{-4\pm 2\sqrt 6}{-4}=\dfrac{2\pm \sqrt 6}{2}$
$x_1=\dfrac{2-\sqrt 6}{2}$
$x_2=\dfrac{2+\sqrt 6}{2}$
Thus, the $x$-intercepts are $\dfrac{2-\sqrt 6}{2}$ and $\dfrac{2+\sqrt 6}{2}$.
