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

Question

Avery Alexander · Accepted Answer

See explanantion

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.

Nathan Axel · Answer

undefined To sketch the graph of $f(x) = -3x^2 + 3x - 2$ using important points, we can follow these steps:

1. Find the vertex: The vertex of a quadratic function in the form $f(x) = ax^2 + bx + c$ is given by the formula $x = -\frac{b}{2a}$. In this case, $a = -3$ and $b = 3$. Plug these values into the formula to find the x-coordinate of the vertex.

2. Once you have the x-coordinate of the vertex, plug it into the function to find the corresponding y-coordinate.

3. Find the y-intercept: The y-intercept occurs when $x = 0$. Plug $x = 0$ into the function to find the y-intercept.

4. Find the x-intercepts: The x-intercepts occur where $f(x) = 0$. Set $f(x)$ equal to zero and solve for $x$.

5. Plot the vertex, y-intercept, and x-intercepts on the coordinate plane.

6. Since the coefficient of $x^2$ is negative, the parabola opens downward. Sketch the parabola passing through these points.

7. Optionally, you can find additional points by choosing other values of $x$ and computing $f(x)$, then plot them to get a more accurate sketch.

Following these steps will help you sketch the graph of the given quadratic function.