# How do you find the vertex and the intercepts for f(x)=5-x^2?

for the x intercepts use the difference of squares to factor. $5 - {x}^{2} = \left(5 - x\right) \left(5 + x\right)$ so x= 5, -5 as the x-intercepts. For the y-int: $\left(5 - {0}^{2}\right) = 5$ the x-value of the vertex is$- \frac{b}{2 a}$ so $x = - \frac{0}{2 - 1} = 0$