Question

What is the vertexof `y= -5x^2 − 3x`?

Answer 1

Vertex: `(frac{-3}{10},frac{9}{20})`

Explanation:

First, use the axis of symmetry formula `(AoS: x = frac{-b}{2a})` to find the x-coordinate of the vertex `(x_{v})` by substituting `-5` for `a` and `-3` for `b`:

`x_{v} = frac{-b}{2a}`

`x_{v} = frac{-(-3)}{2(-5)}`

`x_{v} = frac{-3}{10}`

Then find the y-coordinate of the vertex `(y_{v})` by substituting `frac{-3}{10}` for `x` in the original equation:

`y_{v} = -5x^{2}-3x`

`y_{v} = -5(frac{-3}{10})^{2}-3(frac{-3}{10})`

`y_{v} = -5(frac{9}{100})+frac{9}{10}`

`y_{v} = frac{-45}{100}+frac{90}{100}`

`y_{v} = frac{45}{100}`

`y_{v} = frac{9}{20}`

Finally, express the vertex as an ordered pair:

Vertex: `(x_{v},y_{v}) = (frac{-3}{10},frac{9}{20})`

Answer 2

The vertex is `(-3/10,9/20)` or `(-0.3,0.45)`.

Explanation:

Given:

`y=-5x^2-3x` is a quadratic equation in standard form:

`ax^2+bx-3x`,

where:

`a=-5`, `b=-3`, `c=0`

The vertex of a parabola is its maximum or minimum point. In this case, since `a<0`, the vertex will be the maximum point and the parabola will open downward.

To find the `x`-value of the vertex, use the formula for the axis of symmetry:

`x=(-b)/(2a)`

`x=(-(-3))/(2*(-5))`

`x=3/(-10)`

`x=-3/10`

To find the `y`-value of the vertex, substitute `-3/10` for `x` and solve for `y`.

`y=-5(-3/10)^2-3(-3/10)`

Simplify.

`y=-color(red)cancel(color(black)(5))^1(9/color(red)cancel(color(black)(100))^20)+9/10`

`y=-9/20+9/10`

Multiply `9/10` by `2/2` to get the common denominator `20`.

`y=-9/20+9/10xx2/2`

`y=-9/20+18/20`

`y=9/20`

The vertex is `(-3/10,9/20)` or `(-0.3,0.45)`.

graph{y=-5x^2-3x [-10, 10, -5, 5]}