Question

How do you use the important points to sketch the graph of `f(x)=x^2+2x-1`?

Answer 1

Vertex: `(-1,-2)`

Y-intercept: `(0,-1)`

X-intercepts: `(-1+ sqrt2,0)` and `(-1-sqrt2,0)`

Approximate x-intercepts: `(0.414,0)` and `(-2.414)`.

Additional Points

Point 1: `(3,2)`

Point 2: `(1,2)`

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

Refer to the explanation for the process.

Explanation:

NOTE: This is a very long explanation.

Graph:

`f(x)=x^2+2x-1` is a quadratic equation in standard form:

`f(x)=ax^2+bx+c`,

where:

`a=1`, `b=2`, and `c=-1`

In order to graph a parabola, you need the vertex, and often the x- and y-intercepts, and other points, depending on whether it crosses the x- or y-axes.

Vertex: the maximum or minimum point of the parabola. Since `a>0`, the vertex is the minimum point and the parabola opens upward.

The `x`-value of the vertex is the `x`-value of the axis of symmetry. With a quadratic equation in standard form, the axis of symmetry is:

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

Plug in the known values.

`x=(-2)/(2*1)`

Simplify.

`x=-2/2`

`x=-1`

To find the `y`-value of the vertex, substitute `y` for `f(x)`, and substitute `-1` for `x`.

`y=(-1)^2+2(-1)-1`

Simplify.

`y=1-2-1`

`y=-2`

The vertex is `(-1,-2)`.

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

`y=0^2+2(0)-1`

`y=-1`

The y-intercept is `(0,-1)`.

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

Substitute `0` for `y` and solve for `x` using the quadratic formula.

`x=(-b+-sqrt(b^2-4*a*c))/(2*a)`

Plug in known values.

`x=(-2+-sqrt(2^2-4*1*-1))/(2*1)`

Simplify.

`x=(-2+-sqrt(8))/2`

Prime factorize `8`.

`x=(-2+-sqrt(2xx2xx2))/2`

Simplify.

`x=(-2+-2sqrt2)/2`

Reduce.

`x=(-color(red)cancel(color(black)(2))^1+-color(red)cancel(color(black)(2))^1sqrt2)/color(red)cancel(color(black)(2))^1`

`x=-1+-2sqrt2`

Solutions for `x`.

`x=-1+sqrt2,` `-1-sqrt2`

The x-intercepts are `(-1+ sqrt2,0)` and `(-1-sqrt2,0)`.

Approximate x-intercepts are `(0.414,0)` and `(-2.414)`.

Additional Points.

Point 1

`x=3:` Substitute `-3` for `x` and solve for `y`.

`y=(-3)^2+2(-3)-1`

`y=9-6-1`

`y=2`

Point 1: `(3,2)`

Point 2

`x=1:` Substitute `1` for `x` and solve for `y`.

`y=(-1)^2+2(1)-1`

`y=1+2-1`

`y=2`

Point 2 : `(1,2)`

Summary:

Vertex: `(-1,-2)`

Y-intercept: `(0,-1)`

X-intercepts: `(-1+ sqrt2,0)` and `(-1-sqrt2,0)`

Approximate x-intercepts: `(0.414,0)` and `(-2.414)`.

Additional Points

Point 1: `(3,2)`

Point 2: `(1,2)`

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

graph{y=x^2+2x-1 [-10, 10, -5, 5]}