Question

How do you graph the parabola `y = 2x^2-13x-7` using vertex, intercepts and additional points?

Answer 1

`color(green)(x_("intercpts")=7" and "-1/2)`
`color(green)(y_("intercept") = -7)`
`color(green)(Vertex ->(x,y)=(3 1/4, -225/8))`

Explanation:

Sometimes you are asked to 'sketch' the graph which is very different to doing a full plot.

For the graph (full plot) you would need to build a table of points, mark them and draw a line through them as neatly as you can. This table would have to include the key points. Note that the table should be shown.

For a sketch you just mark and label the key points and roughly draw the line but making sure you make it go through the key points.
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`color(blue)("Determining the key points")`
Given:`" "y=2x^2-13x-7`

The `2x^2` is positive so the graph is of general shape `uu`

`color(brown)(ul("Determine y-intercept"))`

Reading directly off the given equation

This point is at the `x` value 0

`=>y=2x^2-13x-7" "->" "y=0-0-7`

`color(green)(y_("intercept") = -7)`
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`color(brown)(ul("Determine the vertex"))`

There are several approaches for this

A sort of cheat way is to use part of the process of completing the square.

Write as: `" "y=2(x^2-13/2x)-7`

Now apply: `color(green)(x_("vertex")=(-1/2)xx(-13/2) = +13/4 -> 3 1/4)`

Substitute this back into the original equation:

`=>y_("vertex")= 2(13/4)^2-13(13/4)-7`
`= 169/8-169/4-7`
`color(green)(=>y_("vertex")=-225/8)`

`color(green)(Vertex ->(x,y)=(3 1/4, -225/8))`
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`color(brown)(ul("Determine the x intercepts"))`
This will not factorise such that both values are integers.

You can use the formula or complete the square

I chose using the formula.

`y=ax^2+bx+c`

Where a=2; b=-13; c= -7

`x=(-b+-sqrt(b^2-4ac))/(2a) larr" If you can commit this to memory"`

`x=(+13+-sqrt((-13)^2-4(2)(-7)))/(2(2))`

`x=(13+-sqrt(225))/4`

`x=(13+-sqrt(5^2xx3^2))/4`

`x=(13+-15)/4`

`x=28/4" and "-2/4`

`color(green)(x_("intercpts")=7" and "-1/2)`