Question

How do you graph the parabola `y= 2x^2 – 4x – 6` using vertex, intercepts and additional points?

Answer 1

`y=2x^2-4x-6` is a quadratic equation in standard form:

`y=ax^2+bx+c`,

where:

`a=2`, `b=-4`, `c=-6`

Vertex: maximum or minimum point of a parabola

Since `a>0`, the vertex is the minimum point and the parabola opens upward.

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

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

`x=(-(-4))/(2*2)`

`x=4/4`

`x=1`

To find the y-coordinate of the vertex , substitute `1` for `x` in the equation and solve for `y`.

`y=2(1)^2-4(1)-6`

`y=2-4-6`

`y=-8`

The vertex is `(1,-8)`. Plot this point.

Y-intercept: the value of `y` when `x=0`

Substitute `0` for `x` and solve for `y`.

`y=2(0)^2-4(0)-6`

`y=-6`

The y-intercept is `(0,-6)`. Plot this point.

X-intercepts: values of `x` when `y=0`

Substitute `0` for `y` and solve for `x`.

`2x^2-4x-6=0` `larr` I switched sides to get the variable on the left-hand side.

Factor out the common factor `2`.

`2(x^2-2x-3)=0`

Find two numbers that when multiplied equal `-3` and when added equal `-2`. The numbers `1` and `-3` meet the requirement.

`2(x+1)(x-3)=0`

Solve each binomial for `0`.

`x+1=0`

`x=-1`

`x-3=0`

`x=3`

The x-intercepts are `(-1,0)` and `(3,0)`. Plot the points.

Additional points: choose values for `x` and solve for `y`.

Additional point 1

`x=2`

`y=2(2)^2-4(2)-6`

`y=8-8-6`

`y=-6`

Additional point 1 is `(2,-6)`. Plot this point.

Additional point 2

`x=-2`

`y=2(-2)^2-4(-2)-6`

`y=8+8-6`

`y=10`

Additional point 2 is `(-2,10)`. Plot this point.

Additional point 3

`x=4`

`y=2(4)^2-4(4)-6`

`y=32-16-6`

`y=10`

Additional point 3 is `(4,10)`. Plot this point.

Plot the points and sketch a parabola through them. Do not connect the dots.

graph{y=2x^2-4x-6 [-12.21, 10.29, -8.505, 2.745]}