Question

How do you sketch a graph of a quadratic function that has x intercepts 0 and 4?

Answer 1

Refer to the explanation.

Explanation:

If the x-intercepts are `0` and `4`, the quadratic equation must be:

`y=x(x-4)`

Substitute `0` for `y`.

`0=x(x-4)`

`x=0`

`x-4=0`

`x=4`

The points for the x-intercepts are `(0,0)` and `(4,0)`.

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

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

where:

`a=1`, `b=-4`, `c=0`

To graph a quadratic equation, you need the vertex, y-intercept, x-intercepts, and one or two additional points.

You already have the x-intercepts: `(0,0)` and `(4,0)`

Vertex: maximum or minimum point `(x,y)` of the parabola

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

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

`x=(-(-4))/(2*1)`

`x=4/2`

`x=2`

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

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

`y=4-8`

`y=-4`

The vertex is `(2,-4)`.

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

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

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

`y=0`

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

Additional points

Choose values for `x`, substitute them into the equation, and solve for `y`.

Additional Point 1: `x=-1`

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

`y=1+4`

`y=5`

The first additional point is `(-1,5)`.

Additional Point 2: `x=5`

`y=5^2-4(5)`

`y=25-20`

`y=5`

The second additional point is `(5,5)`

Summary

X-intercepts: `(0,0)` and `(4,0)`

Vertex: `(2,-4)`

Y-intercept: `(0,0)`

Additional points: `(-1,5)` and `(5,5)`

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

graph{y=x^2-4x [-10, 10, -5, 5]}