Question

How do you use the important points to sketch the graph of `f(x) = -3x^2 + 3x - 2`?

Answer 1

See explanantion

Explanation:

Once you have determined the general shape of the graph, determined and marked the known points, you sketch the curve free hand as best as you can so that the curve passes through those points.

`color(brown)("The above has actually answered the wording of your question.")`
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However, I am assuming you wish to have more explained!

Given:`" "y=-3x^2+3x-2`

Compare to standard form:`" "y=ax^2+bx+c`

`color(blue)("General shape of the curve")`

The coefficient of `x^2` is -3. The negative property indicates that the general shape of the graph is `nn`

If the coefficient had been +3 then the shape would have been `uu`.

`color(blue)("Vertex"->" maximum as the general shape is "nn`

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`color(blue)("y intercept")`

This can be read directly off the equation in that it is the value of c

`color(blue)(y_("intercept")=-2`

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`color(blue)("Does the curve have an x-intercept")`

Consider the standard form equation solution of

`" "x=(-b+-sqrt(b^2-4ac))/(2a)`

if `b^2-4ac" "` is positive then the graph has at least 1 x-intercept

`a=-3"; "b=+3"; "c=-2`

So `b^2-4ac -> 3^3-4(-3)(-2) = -15`

`color(blue)("There is no "x" intercept")`

`color(brown)("So all of the graph is below the x-axis")`
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`color(blue)("Determine "x_("vertex"))`

Write `color(brown)(y=ax^2+bx+c" as "y=a(x^2+b/ax)+c)`

Write `color(green)(y=-3x^2+3x-2" as "y=-3(x^2-1x)-2)`

`x_("vertex")->(-1/2)xx b/a" " =" "(-1/2)xx(-1) =" " +1/2`

`color(blue)(x_("vertex")=+1/2)`
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`color(blue)("Determine "y_("vertex"))`

Substitute `x=1/2`

`y=-3x^2+3x-2" "->" "y=-3(1/2)^2+3(1/2)-2`

`color(blue)(y_("vertex")=-1 1/4 = -5/4`
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`color(blue)("Vertex" "->(x,y) = (1/2,-5/4)`

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You could always plug a couple of other values for `x` into the equation to give you two more points.