# Quadratic Graphs and Completing the Square

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Quadratic Graphs and Completing the Square
Topic 7.4.3

7.4.3 Topic Quadratic Graphs and Completing the Square
California Standards: 21.0: Students graph quadratic functions and know that their roots are the x-intercepts. 22.0: Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points. What it means for you: You’ll graph quadratic functions by first completing the square of the equation. Key words: quadratic completing the square parabola intercept vertex line of symmetry root

Graph doesn’t cross the x-axis.
Topic 7.4.3 Quadratic Graphs and Completing the Square Graph doesn’t cross the x-axis. If there are no x-intercepts, then it’s impossible to find the vertex by saying that the vertex is halfway between the x-intercepts (like you saw in Topic 7.4.2). But you can use the method of completing the square. This means writing your equation in the form y = (x + k)2 + p.

7.4.3 Topic Quadratic Graphs and Completing the Square
Example 1 Sketch the graph of y = x2 – 6x + 10. Solution You could try to find the x-intercepts by factoring the equation: x2 – 6x + 10 = 0 This time, the left-hand side doesn’t factor. So to find the solutions you could try the quadratic formula with a = 1, b = –6, and c = 10. Solution continues… Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square
Example 1 Sketch the graph of y = x2 – 6x + 10. Solution (continued) Substitute in a = 1, b = –6 and c = 10 Solution continues…

7.4.3 Topic Quadratic Graphs and Completing the Square
Example 1 Sketch the graph of y = x2 – 6x + 10. Solution (continued) However, since you cannot take the square root of a negative number, this tells you that the quadratic function has no real roots — the equation can’t be solved using real numbers. This means that the graph of y = x2 – 6x never crosses the x-axis.

7.4.3 Topic Quadratic Graphs and Completing the Square
But this doesn’t mean that you can’t find the vertex — you just have to use a different method. The trick is to write the equation of the quadratic in the form y = (x + k)2 + p — you need to complete the square. So take the first two terms of the quadratic, and add a number to make a perfect square.

7.4.3 Topic Quadratic Graphs and Completing the Square
Example 1 Sketch the graph of y = x2 – 6x + 10. Solution (continued) Write the equation y = x2 – 6x + 10 in the form y = (x + k)2 + p. x2 – 6x = (x – 3)2 x2 – 6x + 10 = (x – 3)2 – x2 – 6x + 10 = (x – 3)2 + 1 Therefore the function you need to sketch is y = (x – 3)2 + 1. Solution continues…

7.4.3 Topic Quadratic Graphs and Completing the Square
Example 1 Sketch the graph of y = x2 – 6x + 10. Solution (continued) Find the coordinates of the vertex of the graph of y = (x – 3)2 + 1. The minimum value that (x – 3)2 takes is 0 (since a squared number cannot be negative). Therefore the minimum value of y = (x – 3)2 + 1 is = 1. This minimum value occurs at x = 3 (the value for x where (x – 3)2 = 0). So the coordinates of the vertex of the parabola are (3, 1). As before, the line of symmetry passes through the vertex — so the line of symmetry is x = 3.

7.4.3 Topic Quadratic Graphs and Completing the Square
The graph of the quadratic function y = (x + k)2 + p has its vertex at (–k, p). The line of symmetry of the graph is x = –k.

7.4.3 Topic Quadratic Graphs and Completing the Square
Example 1 Sketch the graph of y = x2 – 6x + 10. Solution (continued) The graph of y = x2 – 6x = (x – 3)2 + 1:

7.4.3 Topic Quadratic Graphs and Completing the Square
Example 2 Complete the square for 4x2 – 12x Then find the vertex and line of symmetry of y = 4x2 – 12x + 11. Solution This is a concave-up parabola, since the coefficient of x2 is positive. 4x2 – 12x + 11 Solution continues… Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square
Example 2 Complete the square for 4x2 – 12x Then find the vertex and line of symmetry of y = 4x2 – 12x + 11. Solution (continued) Find the coordinates of the minimum point of the graph of y = 4 x – 3 2 Since the minimum value of is 0, the minimum value of must be = 2. This minimum value occurs at x = . 3 2 and the line of symmetry is x = . So the vertex of the graph of y = 4x2 – 12x + 11 is at , 3 2 As before, put x = 0 to find the y-intercept — this is at y = 11.

Factor out –1 to make completing the square easier.
Topic 7.4.3 Quadratic Graphs and Completing the Square Example 3 Write 4x – x2 – 7 in the form a(x + k)2 + m, and sketch the graph. Solution 4x – x2 – 7 = –x2 + 4x – 7 Factor out –1 to make completing the square easier. = –[x2 – 4x + 7] = –[(x – 2)2 + 3] = –(x – 2)2 – 3 But (x – 2)2 is never negative — the minimum value it takes is 0. So –(x – 2)2 can never be positive, and the maximum value it can take is 0. Solution continues… Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square
Example 3 Write 4x – x2 – 7 in the form a(x + k)2 + m, and sketch the graph. Solution (continued) line of symmetry x = 2 This means that the maximum value of –(x – 2)2 – 3 must be –3, which it takes when x – 2 = 0 — that is, at x = 2. vertex (2, –3) So the vertex of the graph is at (2, –3). y-intercept y = –7 And the line of symmetry is x = 2. As always, find the y-intercept by putting x = 0. This is at y = –7.

7.4.3 Topic Quadratic Graphs and Completing the Square Guided Practice
1. Sketch the graph of the function y = x2 – 12x + 20, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. y-intercept: y = 02 – 12(0) So y-intercept is at (0, 20) x-intercepts: x2 – 12x + 20 = 0 Solving for x gives intercepts at (2, 0) and (10, 0) Vertex and symmetry: x2 – 12x + 20 = (x – 6)2 – = (x – 6)2 – 16 So vertex is at (6, –16) and line of symmetry is x = 6 Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square Guided Practice
2. Sketch the graph of the function y = x2 + 8x + 12, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. y-intercept: y = (0) So y-intercept is at (0, 12) x-intercepts: x2 + 8x + 12 = 0 Solving for x gives intercepts at (–6, 0) and (–2, 0) Vertex and symmetry: x2 + 8x + 12 = (x + 4)2 – = (x + 4)2 – 4 So vertex is at (–4, –4) and line of symmetry is x = –4 Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square Guided Practice
3. Sketch the graph of the function y = x2 – 2x – 3, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. y-intercept: y = 02 – 2(0) – 3 So y-intercept is at (0, –3) x-intercepts: x2 – 2x – 3 = 0 Solving for x gives intercepts at (3, 0) and (–1, 0) Vertex and symmetry: x2 – 2x – 3 = (x – 1)2 – 1 – 3 = (x – 1)2 – 4 So vertex is at (1, –4) and line of symmetry is x = 1 Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square Guided Practice
4. Sketch the graph of the function y = x2 – 4x – 5, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. y-intercept: y = 02 – 4(0) – 5 So y-intercept is at (0, –5) x-intercepts: x2 – 4x – 5 = 0 Solving for x gives intercepts at (5, 0) and (–1, 0) Vertex and symmetry: x2 – 4x – 5 = (x – 2)2 – 4 – 5 = (x – 2)2 – 9 So vertex is at (2, –9) and line of symmetry is x = 2 Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square Guided Practice
5. Sketch the graph of the function y = –x2 – 2x + 3, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. y-intercept: y = –02 – 2(0) + 3 So y-intercept is at (0, 3) x-intercepts: –x2 – 2x + 3 = 0 Solving for x gives intercepts at (–3, 0) and (1, 0) Vertex and symmetry: –x2 – 2x + 3 = –[(x + 1)2 – 1 – 3] = –(x + 1)2 + 4 So vertex is at (–1, 4) and line of symmetry is x = –1 Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square Guided Practice
6. Sketch the graph of the function y = –x2 – x + 6, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. y-intercept: y = –02 – So y-intercept is at (0, 6) x-intercepts: –x2 – x + 6 = 0 Solving for x gives intercepts at (–3, 0) and (2, 0) Vertex and symmetry: –x2 – x + 6 = – x – – 6 = – x So vertex is at and line of symmetry is x = 1 4 2 25 – , Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square Guided Practice
7. Sketch the graph of the function y = –2x2 – 8x + 10, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. y-intercept: y = –2(0)2 – 8(0) So y-intercept is at (0, 10) x-intercepts: –2x2 – 8x + 10 = 0 Solving for x gives intercepts at (1, 0) and (–5, 0) Vertex and symmetry: –2x2 – 8x + 10 = –2[x2 + 4x – 5] = –2[(x + 2)2 – 4 – 5] = –2(x + 2)2 + 18 So vertex is at (–2, 18) and line of symmetry is x = –2 Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square Guided Practice
8. Sketch the graph of the function y = 2x2 + x – 6, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. y-intercept: y = 2(0)2 + 0 – 6 So y-intercept is at (0, –6) x-intercepts: 2x2 + x – 6 = 0 Solving for x gives intercepts at , 0 and (–2, 0) Vertex and symmetry: 2x2 + x – 6 = 2(x x – 3) = 2 x – – 3 = 2 x – So vertex is at and line of symmetry is x = 1 4 2 16 – , – 49 8 3 Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square Guided Practice
9. Sketch the graph of the function y = x2 – 4x + 12, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. y-intercept: y = 02 – 4(0) So y-intercept is at (0, 12) x-intercepts: x2 – 4x + 12 = 0 This equation doesn’t factor and cannot be solved using the quadratic formula. So there are no x-intercepts. Vertex and symmetry: x2 – 4x + 12 = (x – 2)2 – = (x – 2)2 + 8 So vertex is at (2, 8) and line of symmetry is x = 2 Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square Guided Practice
10. Sketch the graph of the function y = 3x2 + 6x + 6, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. y-intercept: y = 3(0)2 + 6(0) + 6 So y-intercept is at (0, 6) x-intercepts: 3x2 + 6x + 6 = 0 This equation doesn’t factor and cannot be solved using the quadratic formula. So there are no x-intercepts. Vertex and symmetry: 3x2 + 6x + 6 = 3(x2 + 2x + 2) = 3{(x + 1)2 – 1 + 2} = 3(x + 1)2 + 3 So vertex is at (–1, 3) and line of symmetry is x = –1 Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square
Independent Practice In Exercises 1–2, use the information that a ball was thrown vertically into the air from a platform m above sea level. The relationship between the height in meters above sea level, h, and the number of seconds since the ball was thrown, t, was found to be h = –5t2 + 6t 3 2 1. After how many seconds did the ball reach its maximum height? 2. What was the ball’s maximum height above sea level? t = s 3 5 h = m 33 10 Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square
Independent Practice 1 8 The first 8 seconds in the flight of a paper airplane can be modeled by the quadratic h = t2 – t + 4, where h is the height in feet and t is the time in seconds. Use this information to answer Exercises 3–4. 3. In the first 8 seconds of its flight, when did the airplane reach its minimum height? 4. What was the minimum height of the plane in the first 8 seconds of its flight? t = 4 s h = 2 ft Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square
Independent Practice A ball is thrown vertically into the air from a platform. The relationship between the ball’s height in meters, h, and the number of seconds, t, since the ball was thrown was found to be h = –5t2 + 10t Use this information to answer Exercises 5–8. 5. After how many seconds did the ball reach its maximum height? 6. What was the maximum height of the ball? 7. At what height was the ball initially thrown? 8. When did the ball hit the ground? t = 1 s h = 20 m h = 15 m t = 3 s Solution follows…

7.4.3 Topic Quadratic Graphs and Completing the Square Round Up
Take a look at Section 7.2 if all this stuff about completing the square seems unfamiliar. Completing the square is a really useful way of graphing quadratics because it gives you the vertex of the graph straightaway.