Quadratic Functions and their Graphs 
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A general quadratic has the form:
y = ax^{2} + bx + c
where a, b and c are constants and a ≠ 0. The simplest of these is:
y = x^{2}
 when a = 1, b = c = 0.
The following observations can be made about this simplest case:
This is sufficient to sketch the function.
In the graph of y = x^{2}:
If the equation is, say, y = 2x^{2}, then the graph will look similar to that of y = x^{2}, but will lie above it. For example, when x = 1 the value of x^{2} is 1, the value of 2x^{2} is 2. The y value of y = 2x^{2} is above that for y = x^{2}. Similarly, for the equation y = x^{2} ⁄ 2, the graph looks similar to that of y = x^{2} but now lies below it. This is illustrated in the figure below.
The next possible choice is a = −1, with the equation y = −x^{2}. In this case the graph of the equation will have the same shape but now, instead of being above the xaxis, it is below. When x=1, the corresponding y value is −1. In a similar way, for differing negative values of a the graphs are below the xaxis.
Quiz 1: The figure below shows a sketch of three quadratics. What can be determined about a, b and c?
Explanation: The curves for y = bx^{2} and y = cx^{2} are both above the xaxis and the former of these is above the latter, so b > c. The curve for y = ax^{2} is below the xaxis so a < 0. Since every positive number is greater than every negative number, it follows that b > c > a.
This type of quadratic is similar to the basic ones in the Introduction section, but with a constant added, i.e. with the general form:
As a simple example of this take the case y = x^{2} + 2. Comparing this with the function y = x^{2}, the only difference is the addition of 2 units. As such:
Similarly, the graph of y = ax^{2} − 2 will be lowered by 2 units.
In the examples considered so far, the axis of symmetry is the yaxis, i.e. the line x = 0. The next possibility is a quadratic which has its axis of symmetry not on the yaxis. An example of this is:
which has the same shape and the same orthogonal axis as y = x^{2} but with the axis of symmetry the line x = 3.
The graph y = (x − 3) is illustrated below:
So far, two separate cases have been discussed: first, a standard quadratic has its orthogonal axis shifted up or down; and second, a standard quadratic has its axis of symmetry shifted left or right. The next step is to consider quadratics that incorporate both shifts.
Example 1: The quadratic y = x^{2} is shifted so that its axis of symmetry is as x = 3 and its orthogonal axis is at y = 2.
Solution: The new curve is symmetric about x = 3 and is shifted up by 2 units, so its equation is y = (x − 3)^{2} + 2.
Solution: The curve crosses the yaxis when x = 0 . Putting this into the equation y = x^{2} + 2, the corresponding value of y is y = (0 − 3)^{2} + 2 = 11, so the curve crosses the yaxis at y = 11.
Solution:
Exercise 1: The curve y = −2x^{2} is shifted so that its axis of symmetry is the line x = −2 and its orthogonal axis is x = 8.
Solution: The equation is:
Solution: The curve cuts the yaxis when x=0. Putting x = 0 into the equation y = −2(x + 2)^{2} + 8, the corresponding y value is:
−2 (0 + 2)^{2} + 8 = −2 (2)^{2} + 8 = − 8 + 8 = 0, i.e. y = 0.
The curve cuts the xaxis when y = 0. In this case, putting the value y = 0 into the equation y = −2(x + 2)^{2} + 8 leads to:
−2 (x + 2)^{2} + 8  =  0 
8  =  2 (x + 2)^{2} 
(x + 2)^{2}  =  4 
(x + 2)  =  ±2 
x  =  −2 ± 2 
so there are two solutions, x = −4 and x = 0.
To summarise: the graph cuts the coordinate axes at two points with coordinates (−4, 0) and (0, 0).
Solution: The curve is sketched below:
Click on questions to reveal their solutions
Exercise 2: For the following curves, write down each curve's equation, find the coordinates of the points where the curve cuts the x and yaxes, and sketch the curve.
Solution: The equation of the shifted curve is
y = (x−7)^{2} + 6
This will cross the yaxis when x = 0, i.e. when:
It does not cross the xaxis since its lowest point is on the orthoganal axis, which is y = 6. A sketch of the graph is shown below:
Solution: This curve will have the same shape as that the curve in part (a), but it is now shifted down rather than up. The equation of the curve is:
This will cross the yaxis when x = 0 and y = (0 − 7)^{2} − 9 = 49 − 9 = 40. It will cross the xaxis when y = 0. Substituting this into the equation gives:
(x−7)^{2} − 9  =  0 
(x−7)^{2}  =  9 
x − 7  =  ±3 
x  =  7 ± 3 
i.e. the curve cuts the xaxis at 4 and 10.
To summarise, the lowest point is on the orthogonal axis at x = 7, y = −9, it crosses the yaxis at y = 40 and it crosses the xaxis at x = 4, x = 10. The curve is sketched below:
Solution: The equation for this curve is:
This will cross the yaxis when x = 0, i.e. at y = −(0 − 7)^{2} + 9 = −49 + 9 = −40. It crosses the xaxis when y = 0, i.e.
−(x − 7)^{2} + 9  =  0 
9  =  (x − 7)^{2} 
x − 7  =  ±3 
x  =  7 ± 3 
which gives x = 4 and x = 10.
To summarise, the curve has its highest point when x = 7 and y = 9, which is the orthogonal axis, it crosses the yaxis at y = −40 and it crosses the xaxis at x = 4 and x = 10. A sketch of the curve is shown below:
Click on questions to reveal their solutions
This section is about sketching general quadratic functions, i.e. ones of the form
ax^{2}+bx+c
The algebraic expression must be rearranged so that the line of symmetry and the orthogonal axis may be determined. The procedure required is completing the square (see module on Quadratics).
Example 2: A quadratic function is given as −2x^{2} + 4x + 16.
Solution: Completing the square:
y = −2x^{2} + 4x + 16  =  −2 (x^{2} − 2x) + 16 
=  −2 [(x − 1)^{2} − 1] + 16  
i.e. y  =  −2 (x − 1)^{2} + 18 
Solution: This is the function y = −2x^{2} moved so that its axis of symmetry is x = 1 and its orthogonal axis is y = 18
Solution: The function is y = −2(x − 1)^{2} + 18. This will cross the xaxis when y = 0, i.e. when:
−2 (x−1)^{2} + 18  =  0 
18  =  2 (x − 1)^{2} 
9  =  (x − 1)^{2} 
taking square roots, x − 1  =  ±3 
x  =  1 ± 3 
=  4 or −2 
Putting x = 0 into the original form of the function y = −2x^{2} + 4x + 16, this gives y = 16, i.e. it crosses the yaxis at y = 16.
Solution:
Here are some exercises for practice:
Exercise 3: Using the method shown in Example 2, sketch each of the following quadratic equations:
Solution: The equation can be rewritten as y = (x + 1)^{2}. A sketch of the function is shown below:
Solution: The function y = 6 − x^{2} is already a complete square and is sketched below:
Solution: On completing the square, the original function y = x^{2} − 6x + 5 becomes y = (x − 3)^{2} − 4.
Solution: On completing the square, the function becomes y = −(x − 2)^{2} + 4. The graph is sketched below:
Solution: On completing the square, the function becomes y = (x + 1)^{2} + 4. The graph is sketched below:
Solution: On completing the square, this function becomes y = −(x + 1)^{2} + 4. This is sketched below:
Click on questions to reveal their solutions
This section ends with a short quiz:
Quiz 2: Which of the following pairs of lines is the axis of symmetry and orthogonal axis respectively of the quadratic function y = − 2x^{2} − 8x?
Explanation: Completing the square on y = −2x^{2} − 8x gives the function
y = −2 (x + 2)^{2} + 8
i.e. the orthogonal axis is y = 8 and the axis of symmetry is x = −2. This is exactly the function examined in Exercise 1, where the full details and sketch may be found.
Each of the following questions relates to the quadratic function y = −x^{2} + 6x + 7.