Question

How do you find the vertex and the intercepts for `y = 2x^2 + 8x + 18`?

Answer 1

Vertex is at `(-2,10)`; Y-intercept: `(0,18)`; No X-ntercept

Explanation:

`y=2x^2+8x+18= 2(x^2+4x)+18=2(x^2+4x+4)+18-8=2(x+2)^2+10:.`Vertex is at `(-2,10)`; Y-intercept: Putting x=0 in the equation we get `y=18 or(0,18)`. X-intercept: Putting y=0 in the equation we get`2(x+2)^2+10=0 or 2(x+2)^2=-10 or (x+2)^2=-5 or x+2= +-sqrt(-5) or x= -2 +- sqrt5 i` graph{2x^2+8x+18 [-40, 40, -20, 20]} [Ans]