064sketchax2plusc

 Year 9 into 10 Step Up Program


 * Sketching y = ±x 2 + k **toc

Standard Parabola
We know //y = x 2 // produces a table of values:

When we graph //y = x 2 //, it produces a parabola. Turning point : (0, 0)

Axis of symmetry : //x = 0//

Y-Intercept (0, 0)

X-Intercept (0, 0)

We also know that y = –x 2 produces an __**inverted**__ parabola.

The question today is:

What happens if we add a value (a constant) after the //x// 2 ?

{By this stage, you should be getting familiar with the shape of the basic parabola so we don't have to use as many entries in the table of values to produce each graph.}

** Question 1 **
Fill out a table of values for //y = x 2 – 2// Then draw the graph on the same axes as //y = x 2 //. State the nature and coordinates of the turning point. Comment about the shape of the graph compared to //y = x 2 //. Estimate the x-intercepts
 * k = //– 2//

Based on this example, what do you think is the rule about how the extra value (+ k) effects the graph? {Check your idea by working through question 2}

** Question 2 **
Use the rule //y = –x 2 + 5// to complete the table Then graph. State the nature and coordinates of the turning point. Comment about the shape of the graph. Estimate the x-intercepts
 * a = //– 1//
 * //k = + 5//

Now write the rule about the effect of the (+ c) on the graph.

** Solution 1 **
Table of values for //y = x 2 – 2//. Now sketch The new graph is __**shifted down**__ by 2

This is called a __**translation**__ ie The graph is __**translated down**__ by 2.

The shape of the parabola has not changed.

__**Minimum turning point**__ at (0, –2)

Axis of Symmetry x = 0

X-Intercepts (approx) (–1.4, 0) and (1.4, 0)

** Solution 2 **
Table of values for y = –x 2 + 5. Now sketch The graph is __**inverted**__ due to the negative sign in front of x 2.

The graph is __**shifted up**__ by 5.

(ie __**translated up**__ by 5)

The shape of the parabola has not changed.

Maximum turning point at (0, 5)

Axis of symmetry : x = 0

X-Intercepts (approx) (–2.2, 0) and (2.2, 0)

Summary
When graphing y = ±x 2 + k.

The value of c causes a __**vertical translation**__ in the parabola.
 * If k > 0 : the parabola will be __**shifted up**__ by k
 * If k < 0 : the parabola will be __**shifted down**__ by the magnitude of k


 * The turning point will be at (0, k)
 * The y-intercept will be at (0, k)
 * The x-intercepts (if they exist) may be read from the graph or calculated by substituting y = 0 and solving for x.

Some things don't change
 * The axis of symmetry is x = 0
 * The shape of the parabola doesn't change (it doesn't get wider or thinner)

** Question 3 **
Use the rule y = x 2 + 3 to complete the table and then sketch State the nature and coordinates of the turning points. State the equation of the axis of symmetry State the x-intercepts.

** Question 4 **
Sketch y = –x 2 – 1

Try to sketch this without doing a table of values. Decide where the turning point will be and sketch from there.

State the nature and location of the turning point State the equation of the axis of symmetry State the x-intercepts

** Solution 3 **
Table of values for y = x 2 + 3 Now sketch The graph is __**upright**__ and __**shifted up**__ by 3 (or __**translated up**__ by 3)

__**Minimum**__ turning point at (0, 3)

Axis of symmetry is x = 0

X-Intercepts (none)

** Solution 4 **
Sketch y = –x 2 – 1

{Graph will be inverted and shifted down by 1 so turning point is at (0, –1)}

Graph is inverted and shifted down by 1

Maximum turning point at (0, –1)

Axis of symmetry : x = 0

X-Intercepts (none)



** Compare to y = mx + c **
The c value in y = x 2 + c can be compared to the c value in the equation of a straight line : y = mx + c.

In both the straight line and the parabola (in this form), the y-intercept will be (0, c)

In both the straight line and the parabola, the c value shifts the standard graph up or down by a distance of c from the origin.

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