Question

How do you draw the graph of `y=3x^2-7`?

Answer 1

Note that `y = 3x^2 -7` is a parabola (which opens upward) so we will need to calculate a few sample points and then draw a parabolic curve through those points.

For example you might calculate `f(x)` for all integer values in the range `[--4,+4]` (note symmetry about the Y-axis reduces the number of actual calculations).
`(-4,41)` and `(+4,41)`
`(-3,20)` and `(+3,20)`
`(-2, 5)` and `(+2,5)`
`(-1,-4)` and `(+1,-4)`
`(0,-7)`

Mark these point on some graph paper and joint them in a smooth curve that should look something like:
graph{3x^2-7 [-9.71, 10.29, -7.16, 2.84]}

Answer 2

Refer to explanation.

Explanation:

Given:

`y=3x^2-7` is a quadratic equation in standard form:

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

where:

`a=3`, `b=0`, `c=-7`

To graph a parabola you need to determine the following points:

axis of symmetry, vertex, y-intercept (if applicable), x-intercepts (if applicable), and additional points if needed.

Axis of symmetry: vertical line that divides the parabola into two equal halves

Vertex: the maximum or minimum point of the parabola.

The formula for determining the axis of symmetry, which is also the `x`-coordinate of the vertex is:

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

`x=0/(2*3)`

`x=0`

To determine the `y`-coordinate by substituting `0` for `x` and solving for `y`.

`y=3(0)^2-7`

`y=-7`

The vertex is `(0,-7)`

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

In this particular case, the y-intercept is the same as the vertex, `(0,-7)`.

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

`0=3x-7`

Add `7` to both sides.

`7=3x^2`

Divide both sides by `3`.

`7/3=x^2`

`+-sqrt(7/3)=x`

`x=-sqrt(7/3),` `sqrt(7/3)`

The x-intercepts are `(-sqrt(7/3),0)` and `(sqrt(7/3),0)`.

The approximate x-intercepts are `(-1.528,0)` and `(1.528,0)`.

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

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

graph{y=3x^2-7 [-9.88, 10.12, -7.88, 2.12]}