Question

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

Answer 1

See below

Explanation:

for the x intercepts use the difference of squares to factor. `5-x^2=(5-x)(5+x)` so x= 5, -5 as the x-intercepts. For the y-int: `(5-0^2)=5` the x-value of the vertex is`-b/(2a)` so `x=-0/(2-1)=0`
Then you insert the x-value of the vertex to get the y-value of the vertex. So the vertex would also be the y-int (0,5)