How to graph a parabola #x=(y^2) - 4y + 3#?

2 Answers
Jul 9, 2015

You prepare a chart of #x# and #y# values and plot the points.

Explanation:

#x= y^2 -4y +3#

Note that #x# is the dependent variable and #y# is the independent variable.

Step 1. Prepare a chart.

Try an interval from #y = -5# to #y = 5#, and calculate the corresponding values of #x#.

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Step 2. Plot these points.

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Step 3. Add points to make the plot symmetrical.

We need some extra points on the top portion of the graph.

Let's extend our table to #y = 10#.

Here's the extended portion of the table.

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Add these extra points to the plot.

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And we have our graph.

Jul 9, 2015

If the question is to sketch the parabola, you plot the vertex and the #x#- and #y#-intercepts, and draw a smooth line through them.

Explanation:

Warning! This is a long answer.

#x= y^2 -4y +3#

Note that #x# is the dependent variable and #y# is the independent variable.

We are going to get a sideways parabola.

Step 1. Define your variables.

The standard form for the equation of this parabola is

#x = ay^2 +by +c#

So

#a= 1#, #b = -4#, and #c = 3#

Step 2. Calculate and plot the vertex.

The vertex of the curve is given by

#y = -b/(2a) = -(-4)/(2×1) = -(-2) = 2#

Calculate the x-coordinate of the vertex.

#x= y^2 -4y +3 = (2)^2 -4(2) + 3 = 4 -8 + 3 = -1#

So the vertex is at (#-1,2#).

Plot your vertex point.

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Step 3. Find the direction of the opening.

The parabola will be a sideways U opening either to the right (#⊂#) or to the left (#⊃#).

Since the coefficient #a# is positive, the parabola opens in the positive direction (to the right).

Step 4. (optional) Draw the parabola's axis of symmetry.

A parabola's axis of symmetry is a line that runs through its middle and divides it in half.

For a quadratic of the form #x = ay^2 +by +c#, the axis is a line that passes through the vertex and is parallel to the #y# axis.

For our parabola, the axis is the line #y = 2.

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It's not part of the parabola itself, but lightly marking this line on your graph can help you see how the parabola curves symmetrically.

Step 5. Calculate and plot the #x#-intercept.

#x= y^2 -4y +3 = (0)^2 -4(0)+3 = 0 – 0 + 3 = 3#

The #x#-intercept is at (#3,0#). Plot this point.

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Step 6. Calculate and plot any #y#-intercepts.

#f(y) = y^2 -4y +3 = 0#
#(y-3)(y-1) = 0#
#y-3 = 0# or #y-1 = 0#
#y = 3# or #y = 1#

The x-intercepts are at (#0,3#) and (#0,1#).

Add these points to the graph.

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Step 7. Add any extra points to the graph.

The #x#-intercept at (#3,0#) is 2 units below the axis. There should be a corresponding point 2 units above the axis at (#3,4#).

Plot this point.

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Step 7. Draw a smooth parabola passing through all the points.

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