Polynomials: Graphing Polynomials. By Mr Porter

Assumed Knowledge: It is assumed that students can fully factorise polynomials. Also, student should know how to * factorise quadratic expression, ax2 + bx + c * determine if a quadratic is positive or negative DEFINITE (using the discriminate ∆ = b2 – 4ac ). * Solve simple linear equations, mx + b = 0 Student should be familiar with the coordinate number plane. Quadrant 1 Q1 Quadrant 2 Q2 Quadrant 3 Q3 Quadrant 4 Q4 X-axis Y-axis Note: Y-axis is also the polynomial P(x).

Example 1: Graph the factorised polynomial, P(x) = x ( x – 3 )( x + 2 ). Step 1: Find the x-intercepts, take each factor and set them equal to zero and solve. x = 0 x – 3 = 0  x = 3 x +2 = 0  x = -2 Step 5: Start drawing the polynomial, the curve must pass through each of the x-intercepts and the vertical intercept P(x) = x ( x – 3 ) ( x + 2 ) Step 2: Draw a simple number plane, mark and label the three solution x = -2 , x = 0 and x = 3. P(x) X-axis • 3 -2 Step 3: Starting point! Take each of the x-terms and multiply them. x x x x x = 1x3 Note, it is POSITIVE. We start in the first quadrant, Q1. x These points are guesses! If you want accuracy, evaluate P(-1) and P(1.5) Step 4: Find the vertical axis intercept (it is the constant term) Note x = (x – 0). So, take each of the constant terms and multiply: (0) x (-3) x (+2) = 0 The vertical intercept is 0. Remember, it is a sketch!

Example 2: Graph the factorised polynomial, P(x) = x(x + 1)( x + 3 )( x – 4). Step 1: Find the x-intercepts, take each factor and set them equal to zero and solve. x = 0 x + 1 = 0  x = -1 x + 3 = 0  x = -3 x – 4= 0  x = 4 Step 5: Start drawing the polynomial, the curve must pass through each of the x-intercepts and the vertical intercept P(x) = x(x + 1) ( x + 3 ) ( x – 4) Step 2: Draw a simple number plane, mark and label the four solution x = -3 , x = -1, x = 0 and x = 4. P(x) X-axis • 4 -3 -1 Step 3: Starting point! Take each of the x-terms and multiply them. x x x x x x x = +1x4 Note, it is POSITIVE. We start in the 1st quadrant, Q1. x Step 4: Find the vertical axis intercept (it is the constant term) Note x = (x – 0). So, take each of the constant terms and multiply: (0) x (+1) x (+3) x (-4) = 0 The vertical intercept is 0. Remember, it is a sketch!

Example 3: Graph the factorised polynomial, P(x) = (x + 1)( x + 3 )( 2 – x ). Step 1: Find the x-intercepts, take each factor and set them equal to zero and solve. x + 1 = 0  x = -1 x + 3 = 0  x = -3 2 – x = 0  x = 2 Step 5: Start drawing the polynomial, the curve must pass through each of the x-intercepts and the vertical intercept P(x) = (x + 1) ( x + 3 ) ( 2 – x) Step 2: Draw a simple number plane, mark and label the three solution x = -3 , x = -1 and x = 2. P(x) X-axis • 2 -3 -1 Step 4: Find the vertical axis intercept (it is the constant term) So, take each of the constant terms and multiply: (+1) x (+3) x (+2) = +6 The vertical intercept is +6. 6 • Step 3: Starting point! Take each of the x-terms and multiply them. x x x x -x = -1x3 Note, it is NEGATIVE. We start in the 4th quadrant, Q4. x Remember, it is a sketch!

Example 4: Graph the factorised polynomial, P(x) = (x + 1)( x + 3 )( x – 4)(5 – x). Step 1: Find the x-intercepts, take each factor and set them equal to zero and solve. x + 1 = 0  x = -1 x + 3 = 0  x = -3 x – 4 = 0  x = 4 5 – x = 0  x = 5 Step 5: Start drawing the polynomial, the curve must pass through each of the x-intercepts and the vertical intercept P(x) = (x + 1) ( x + 3 ) ( x – 4) (5 – x) Step 2: Draw a simple number plane, mark and label the four solution x = -3 , x = -1, x = 4 and x = 5. P(x) X-axis • 4 -3 -1 5 Remember, it is a sketch! Step 3: Starting point! Take each of the x-terms and multiply them. x x x x x x -x = -1x4 Note, it is NEGATIVE. We start in the 4th quadrant, Q4. x Step 4: Find the vertical axis intercept (it is the constant term) So, take each of the constant terms and multiply: (+1) x (+3) x (-4) x (+5)= -60 The vertical intercept is -60. -60 •

Example 5: Sketch the polynomial, P(x) = x3 – 4x2 + 4x. Step 5: Find the vertical axis intercept (it is the constant term) . In this case, it is the origin, (0, 0) The vertical intercept is 0. Step 6: Start drawing the polynomial, the curve must pass through each of the x-intercepts and the vertical intercept P(x) = x3 – 4x2 + 4x Step 1: Factorise the polynomial. P(x) = x3 – 4x2 + 4x P(x) = x(x – 2) (x – 2) Step 3: Draw a simple number plane, mark and label the two solution x = 0 and x = 2. P(x) X-axis • 2 Step 2: Find the x-intercepts, take each factor and set them equal to zero and solve. x = 0 x – 2 = 0  x = 2 Step 4: Starting point! Since the leading coefficient is +1 of x3 Note, it is POSITIVE. We start in the 1st quadrant, Q1. x Remember, it is a sketch! The two values at x = 2, means that the polynomial has a turning point at x = 2.

Example 6: Sketch the polynomial, P(x) = 2x3 + 7x2 + 4x – 4. Step 5: Find the vertical axis intercept (it is the constant term) . In this case, it is -4 The vertical intercept is -4. • -4 Step 6: Start drawing the polynomial, the curve must pass through each of the x-intercepts and the vertical intercept P(x) = 2x3 + 7x2 + 4x – 4 Step 1: Factorise the polynomial. P(x) = 2x3 + 7x2 + 4x – 4 P(x) = (2x – 1) (x + 2) (x + 2) Step 3: Draw a simple number plane, mark and label the two solution x = 1/2 and x = -2. P(x) X-axis • -2 1/2 Step 2: Find the x-intercepts, take each factor and set them equal to zero and solve. 2x – 1 = 0  x = 1/2 x + 2 = 0  x = -2 Step 4: Starting point! Since the leading coefficient is +2 of x3 Note, it is POSITIVE. We start in the 1st quadrant, Q1. x Remember, it is a sketch! The two values at x = -2, means that the polynomial has a turning point at x = -2.

Example Polynomials graphs generated using a FX-Graph application. P(x) = x(x + 1) ( x + 3 ) ( x – 4) P(x) = x ( x – 3 ) ( x + 2 ) Ex 1 Ex 2 P(x) = (x + 1) ( x + 3 ) ( 2 – x) P(x) = (x + 1) ( x + 3 ) ( x – 4) (5 – x) Ex 3 Ex 4