How do you find the vertex and the intercepts for `y=2x^2 + 8x + 5`?

Since the quadratic equation is given in standard form, we can find the vertex and roots using the vertex and quadratic formula.

Standard form is `ax^2+bx+c`. In this case:

`a=2`

`b=8`

`c=5`

The vertex formula for the `x`-coordinate is `-\frac{b}{2a}`

Plugging in yields:

`-\frac{8}{2\cdot 2}`

`=-2`

To find the `y`-coordinate, plug in the `x`-coordinate (`-2`) into the original quadratic equation, in place of `x`:

`2x^2+8x+5`

`\Rightarrow 2(-2)^2+8(-2)+5`

`\Rightarrow 2(4)-16+5`

`\Rightarrow 8-16+5`

`\Rightarrow -3`

Therefore, the vertex is `(-2,-3)`

Now, to find the `x`-intercepts, plug the corresponding `a`, `b`, and `c` values into the quadratic formula

`x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}`

`x=\frac{-8\pm\sqrt{8^2-4(2)(5)}}{2(2)}`

`x=\frac{-8\pm\sqrt{64-40}}{4}`

`x=\frac{-8\pm\sqrt{24}}{4}`

`x=\frac{-8\pm 2\sqrt{6}}{4}`

Splitting `x` up into the plus and minus values:

`x=\frac{-8+2\sqrt{6}}{4},\qquad\qquad\qquad x=\frac{-8-2\sqrt{6}}{4}`

`x=-3.2247448713916,\qquad\qquad\qquadx=-0.77525512860841`

Those are the `x`intercepts.

The `y`-intercept is simply `(0,c)`, or `(0,5)` in this case.

Bada-bing, Bada-boom.