5 Solving from Vertex Form One characteristic of numbers to be cautious about is that when we square a real number, squaring will effectively remove any negative sign.x2 = 25(5)2 = 25 (–5)2 = 25x = 5 x = –5x =We know that 5 squared is 25.Negative 5 squared is also 25.Both must be given as answers to this equation.When using a square root, we must use the plus/minus symbol to represent both answers.
6 Solving from Vertex Form When we write a square root, we are referring to the positive solution, which is called the principal square root. Therefore, = 5.Because the principal square root only accounts for positive solutions we lose possible negative results, so we must use a plus/minus symbol () to show that there are two possible answers.Using the square root and plus/minus is an example of the square root property.
7 Solving from Vertex Form With this property in mind, we are going to look at two basic ways to solve quadratic equations: the square root property and completing the square. First, we will use the square root property to solve quadratics when they are given in vertex form.
9 Example 1 – Solving quadratic equations using the square root property Solve 10 = 2(x – 4)2 – 8.Solution:10 = 2(x – 4)2 – 818 = 2(x – 4)29 = (x – 4)2Isolate the squared variable expression.
10 Example 1 – Solution 3 = x – 4 3 = x – 4 or –3 = x – 4 cont’d 3 = x – 43 = x – 4 or –3 = x – 47 = x or 1 = x10 ≟ 2(7 – 4)2 – 810 = 10Use the square root property. Don’t forget the plus/minus symbol.Rewrite as two equations andsolve.Check both answers.
11 Example 1 – Solution 10 ≟ 2(1 – 4)2 – 8 10 = 10 cont’d10 ≟ 2(1 – 4)2 – 810 = 10x = 7 and x = 1 are both valid answers to this equation.Both answers work.
13 Completing the SquareIf a quadratic equation has both a second-degree term and a first-degree term (ax2 + bx + c), the square root property cannot be easily used to solve.If we try to get the squared variable expression by itself, we will have a variable term in the way of the square root. To handle this problem, we use a technique called completing the square.Completing the square will transform the equation so that it has a perfect square that can be solved using the square root property.
14 Completing the SquareIn general, taking half of the coefficient of the first-degree term and squaring it will give us a constant that will make these expressions a perfect square trinomial.x2 + bxThe factored form will include half of b as one of the terms in the binomial.The constant that will completethe square:
15 Completing the SquareWe will start the process of completing the square by simplifying one side of an equation.This process is easiest when the coefficient of the squared term is 1. If this coefficient is not 1, we will divide both sides of the equation by the coefficient to make it 1.
16 Example 4 – Solving quadratic equations by completing the square Solve by completing the square of 5x2 + 30x – 35 = 0.Solution:Step 1: Isolate the variable terms on one side of the equation.To isolate the variable terms, we move the constant term to the other side of the equation.5x2 + 30x – 35 = 05x2 + 30x = 35
17 Example 4 – Solutioncont’dStep 2: If the coefficient of x2 is not 1, divide both sides of the equation by the coefficient of x2.Divide both sides by the coefficient, 5.x2 + 6x = 7Step 3: Take half of the coefficient of x, then square it. Add this number to both sides of the equation.
18 Example 4 – Solutioncont’dTo complete the square on the left side of the equation, we add a number that will make the resulting trinomial a perfect square trinomial.To find this number, we take half of the coefficient of x and square it. This number is then added to both sides of the equation. In this equation, the coefficient of x is 6, so we get
19 Example 4 – Solution Add this constant to both sides of the equation. cont’dAdd this constant to both sides of the equation.x2 + 6x + 9 = 7 + 9x2 + 6x + 9 = 16Step 4: Factor the quadratic into the square of a binomial The left side will factor as a perfect square.(x + 3)2 = 16
20 Example 4 – Solution Step 5: Solve using the square root property. cont’dStep 5: Solve using the square root property.(x + 3)2 = 16x + 3 = 4x + 3 = 4 x + 3 = –4x = x = –7Use the square root property.Rewrite as two equationsand solve.
21 Example 4 – Solutioncont’dStep 6: Check your answers in the original equation.5(1)2 + 30(1) – 35 ≟ 0( – 35) ≟ 00 = 0
25 Converting to Vertex Form To put a quadratic function into vertex form, use the technique of completing the square.The process is very similar, but instead of adding the constant that will complete the square to both sides, you will add and subtract the constant on one side.
26 Example 6 – Converting a quadratic function into vertex form Convert to vertex form.a. f (x) = 2x2 + 20x – 6 b. g(x) = 5x2 + 3x + 20Solution:a. Converting a function to vertex form will be similar to completing the square, as we did earlier. Because we are not solving, we will keep everything on one side of the equation, leaving the function notation alone.f (x) = 2x2 + 20x – 6f (x) = (2x2 + 20x) – 6Group the variable terms.Factor out the coefficientof the squared term.
27 Example 6 – Solution f (x) = 2(x2 + 10x) – 6 cont’df (x) = 2(x2 + 10x) – 6Now we want to complete the square of the variable term in parentheses. Find the constant that will complete the square.52 = 25Add and subtract 25 inside the parentheses. Then take out the subtracted 25, keeping the multiplication by Finally, you can simplify and factor.f (x) = 2(x2 + 10x + 25 – 25) – 6Add and subtract 25 inside the parentheses.
28 Example 6 – Solution f (x) = 2(x2 + 10x + 25) – 2(25) – 6 cont’df (x) = 2(x2 + 10x + 25) – 2(25) – 6f (x) = 2(x2 + 10x + 25) – 50 – 6f (x) = 2(x + 5)2 – 56b. g(x) = 5x2 + 3x + 20g(x) = (5x2 + 3x) + 20Bring out the subtracted 25, keeping the multiply by 2 from the parentheses.Simplify and factor.Group the variable terms.Factor out the coefficient of the squared term.
29 Example 6 – Solution Find the constant that will complete the square. cont’dFind the constant that will complete the square.Add and subtractinside the parentheses.Bring out the subtracted , keeping the multiply by 5 from the parentheses.
32 Graphing from Vertex Form with x-Intercepts Now that we know how to solve a quadratic in vertex form, we can find the horizontal intercepts of a quadratic in vertex form, and these can be part of our graphs.
33 Step 1: Determine whether the graph opens up or down. Example 7 – Graphing a quadratic function in vertex form including horizontal interceptsSketch the graph of f (x) = –1.5(x – 2.5) Give the domain and range of the function.Solution:Step 1: Determine whether the graph opens up or down.The value of a is –1.5. Because a is negative, the graph will open downward.Step 2: Find the vertex and the equation for the axis of symmetry.Because the quadratic is given in vertex form, the vertex is (2.5, ).The axis of symmetry is the vertical line through the vertex x = 2.5.
34 Example 7 – Solution Step 3: Find the vertical intercept. cont’dStep 3: Find the vertical intercept.To find the vertical intercept, we make the input variable zero.f (x) = –1.5(x – 2.5)f (0) = –1.5(0 – 2.5)f (0) = –1.5(–2.5)f (0) = –1.5(6.25)f (0) = 36Therefore, the vertical intercept is (0, 36).
35 Example 7 – Solution Step 4: Find the horizontal intercepts (if any). cont’dStep 4: Find the horizontal intercepts (if any).Horizontal intercepts occur when the output variable is equal to zero, so substitute zero for the output variable and solve.30.25 = (x – 2.5)2Isolate the squared variable expression.Use the square root property.
36 Example 7 – Solution 5.5 = x – 2.5 5.5 = x – 2.5 –5.5 = x – 2.5 cont’d 5.5 = x – 2.55.5 = x – –5.5 = x – 2.58 = x –3 = xTherefore, we have the horizontal intercepts (–3, 0) and (8, 0).Write two equations and solve.
37 Example 7 – Solutioncont’dStep 5: Plot the points you found in steps 2 through 4. Plot their symmetric points and sketch the graph. (Find an additional pair of symmetric points if needed.)The vertical intercept (0, 36) is 2.5 units to the left of the axis of symmetry x = 2.5, so its symmetric point will be units to the right of the axis of symmetry at (5, 36).
38 Example 7 – Solutioncont’dConnect the points with a smooth curve. The input variable x can be any real number without causing the function to be undefined, so the domain of the function is all real numbers or ( , ).The graph decreases to negative infinity, and the highest output, y = , is at the vertex so the range is ( , ] or y