The aim of this powerpoint is to enable you to quickly and easily draw the graphs of parabolas when given their equations. It will show you how to determine whether or not a parabola will intersect the x-axis determine the turning point of a parabola organise the appearance of the parabola’s equation to give key information about its features.

The equation of a parabola The general equation of a parabola is any equation of the format y = ax 2 + bx + c where a, b, c are constants. b and c can equal anything including zero, in which case the equation may be missing the second and/or the third term. But a can NEVER be zero, otherwise the x 2 term disappears and you would no longer have a parabola, but instead, y = bx + c which is a straight line!

Here are some equations of parabolas abcEquation 253y = 2x 2 + 5x – 4y = x 2 + x – 4 307y = 3x ½– 80y = ½x 2 – 8x 100y = x 2 – 219y = – 2x 2 + x + 9 – 104y = 4 – x 2 Note – the red one is the basic parabola, y = x 2. Note also sometimes the terms are mixed up – be on the lookout for this! y = 2x – 7 – x 2

Special features – LEARN THESE! If a > 0 it’s upright ; if a < 0 it’s inverted If b is zero (i.e. no “x”-term) then it’s symmetrical about the y-axis. No sideways movement. If c is zero (i.e. no constant on the end) then it passes through the origin. The origin is both an x and y intercept. The x-value of the VERTEX (turning point) is the MEAN of the x-values of the two x intercepts. Now check the table on the previous page and identify their features!

abcEquation 253y = 2x 2 + 5x + 3 UPRIGHT 11– 4y = x 2 + x – 4 UPRIGHT 307y = 3x UPRIGHT, SYMMETRICAL ½– 80y = ½x 2 – 8x UPRIGHT, GOES THROUGH ORIGIN 100y = x 2 UPRIGHT, THROUGH ORIGIN, SYMMETRICAL – 219y = – 2x 2 + x + 9 INVERTED – 104y = 4 – x 2 INVERTED, SYMMETRICAL

Drawing parabolas on the TI Example To draw the graph of y = x 2 – 4x + 3 STEP 1 Select Y= and enter equation STEP 2 Select WINDOW & enter Xmin, Xmax values only. Ignore y values. STEP 3 Select ZOOM 0 (i.e. ZOOMFIT)

Drawing parabolas on the TI – cont’d Note we have set WINDOW X- range as – 5 to 5. A better graph can be drawn by narrowing this a bit. Try adjusting WINDOW by changing Xmin to 0 and Xmax to 4. Then ZOOM 0. This is a much better representation as it more clearly shows x-intercepts A, B and the location of the turning point T AB T 0 4 There is a fair bit of trial and error (mostly error!) associated with setting the WINDOW. Begin with -5 to 5, hit ZOOM 0 and then experiment by adjusting your values to get a better picture.

Using the TABLE facility on the TI This is a very handy facility as it lets you see the coordinates of all points lying on your parabola. It is especially useful if you have to draw a graph by hand and it’s not feasible just to copy the graph off the screen. STEP 1 Select Y= and enter equation as before STEP 2 Select TBLSET. This lets you choose where you start and what the x-values go up by (usually 1’s). Set TblStart = 0 Set ΔTbl = 1

STEP 3 The TABLE facility on the TI – cont’d Select TABLE Note the first x value is 0 (we set this up with TblStart) and the x’s jump by 1’s (we set this up with ΔTbl). All these points (0, 3), (1, 0), (2, -1) etc lie on your parabola Note also that you can use the up and down arrow keys to SCROLL up and down this table to see other points

Using TABLE to locate key points (0, 3) is the y-intercept (1, 0) and (3, 0) are the x-intercepts We can also determine the vertex from this table. (2, -1) is the vertex Note how the y values are symmetrical about the value –1, i.e. reading down the y-values notice that they follow a pattern 3, 0, -1, 0, 3…. If this happens, the CENTRE of the pattern is on the axis of symmetry going through the middle of the parabola, and so is the turning point. SO, the Vertex is (2, – 1). See next slide for the visual

0 x y This vertical line, x = 2, is called the AXIS OF SYMMETRY. It splits the parabola in half and contains the Turning Point y-intercept (0, 3) x-intercepts (1,0) & (3,0) Vertex (called a “Minimum Turning Point ” when it is the lowest point) (2, – 1)

A note about the y-intercept For the parabola y = ax 2 + bx + c, the y-intercept is obtained by substituting x = 0 into the equation. So, replacing x with 0, we obtain y = a (0) 2 + b(0) + c i.e. y = c So the y-intercept of the parabola y = ax 2 + bx + c is equal to c. Compare with straight lines y = mx + c which also have a y-intercept equal to c ! c

Now try graphing the parabolas in the table on Slide #4.

So far we have…. Looked at parabolas whose equations are of the form y = ax 2 + bx + c. This is called GENERAL FORM Investigated some facilities available on the TI relating to drawing graphs and obtaining tables of values. Used TABLE to identify intercepts and turning points. Now we will consider how to draw parabolas using algebra, rather than relying on the graphics.

Factorised form for the equation of a parabola. Also called x-intercept form. In previous slides we looked at the equation of the parabola y = x 2 – 4x + 3. (called GENERAL FORM) If we FACTORISE the right hand side, we obtain y = (x – 1)(x – 3) This is known as the FACTORISED FORM for the equation of a parabola and its major use is for determining x – intercepts. If you’re not using a graphics, pinpointing the x – intercepts will enable you to graph the parabola quickly and easily!

Example Convert y = x 2 – 4x – 5 to factorised form, and determine its x and y intercepts, turning point, equation of axis of symmetry. STEP 1 Finding the y-intercept The y-intercept is equal to c and so is – 5 STEP 2 Finding the x-intercepts Factorising the right hand side x 2 – 4x – 5 = (x – 5)(x + 1) so our parabola’s equation becomes y = (x – 5)(x + 1)

STEP 2 Finding the x-intercepts It is important that you know: x-intercepts lie on the x-axis and therefore have no height, i.e. y = 0 y = (x – 5)(x + 1) So when we make y = 0, this means we make (x – 5)(x + 1) = 0 and solve this quadratic. The solutions will be the x-intercepts. (x – 5)(x + 1) = 0 So x – 5 = 0 OR x + 1 = 0 Which gives x = 5 and x = – 1 as our x-intercepts

STEP 3 Plotting the intercepts So far we know the y-intercept is (0, – 5) and x-intercepts are (5, 0) and (– 1, 0). We plot these below:   

   STEP 4 Plotting the axis of symmetry The axis of symmetry cuts the parabola in half. It cuts through the x- axis at the point halfway between the two x-intercepts. This point is just the mean of – 1 and 5, i.e. (– 1 + 5) ÷ 2 which is 2.  We can now draw the axis of symmetry.

STEP 5 Plotting the Turning Point The turning point lies on the axis of symmetry and so has an x-value equal to 2. It now remains for us to find the y- value for the turning point. This is done by substituting x = 2 into the parabola’s equation y = x 2 – 4x – 5 Substituting x = 2 y = 2 2 – 4(2) – 5 = – 9 Turning pt is (2, – 9)     

STEP 6 Joining the dots!     It’s now just a question of joining the dots. Remember the parabola is symmetrical about the vertical dotted line. You can also plot other points for greater accuracy.

Example (shorter than the last one!) Use algebra to determine y-intercept, x-intercepts, turning point and axis of symmetry of the parabola y = 4 – 2x – 2x 2 STEP 1 Y-INTER Rearrange into normal quadratic form y = – 2x 2 – 2x + 4 STEP 2 X-INTERS Note that the a value is – 2, and as it’s negative the parabola will be upside-down (inverted). Take out a common factor, with the negative. y = – 2x 2 – 2x + 4 Y – INTERCEPT = (0, 4) y = – 2(x 2 + x – 2)

Factorise the brackets y = – 2(x + 2)(x – 1) X – INTERCEPTS = ( – 2, 0) AND (1, 0) STEP 3 TURNING POINT AXIS OF SYMM Mean of the two x-intercepts is – ½. This is the x – value of the turning point. So the turning point is ( – ½, ? ) To find the y-value (?), substitute x = – ½ into original equation y = 4 – 2x – 2x 2 TURNING PT IS (– ½, 4 ½) AND THE AXIS OF SYMMETRY IS x = – ½

x inter (1, 0) x inter (–2, 0) y inter (1, 0) Turn pt (– ½, 4 ½ )

We have investigated two formats for equations of parabolas… General form y = ax 2 + bx + c Also known as expanded form, this quickly tells you the y-intercept is c. The sign of a tells you whether it’s upright or inverted. Factorised form y = a(x – α)(x – β) This is useful as it tells you the x- intercepts are at x = α and x = β.

We have also worked out that the TURNING POINT has as its x-coordinate the MEAN (average) of the two x-intercepts….. …..and that the AXIS OF SYMMETRY is the vertical line that splits the parabola in half. The axis of symmetry contains the TURNING POINT and it crosses the x-axis halfway between the two x-intercepts…..

We look at the third version of the equation of the parabola – the VERTEX (turning point) format. Investigation…. (2) On your graphics, plot y = (x – 2) 2 – 1. (1) Expand and simplify (x – 2) 2 – 1. (3) Use your graphics to determine the vertex of y = (x – 2) 2 – 1.

Solution…. (1) Expand and simplify (x – 2) 2 – 1. (x – 2) 2 – 1 = x 2 – 4x + 4 – 1 = x 2 – 4x + 3 (2) Enter Y 1 = (X – 2) 2 – 1 Adjust window

Move cursor a little to the RIGHT of the vertex. (3) Use your graphics to determine the vertex of y = (x – 2) 2 – 1. Select 2 nd TRACE 3 Move cursor a little to the LEFT of the vertex. Hit ENTER Hit ENTER Hit ENTER

The Vertex is therefore at (2, – 1)