Quadratic functions Elementary Functions In the last lecture we - PowerPoint PPT Presentation

Quadratic functions Elementary Functions In the last lecture we studied polynomials of simple form f ( x ) = mx + b. Now we move on to a more interesting case, polynomials of degree 2, the Part 2, Polynomials quadratic polynomials. Lecture 2.1a, Quadratic Functions Quadratic functions have form f ( x ) = ax 2 + bx + c where a, b, c are real Dr. Ken W. Smith numbers and we will assume a � = 0 . Sam Houston State University The graph of a quadratic polynomial is a parabola. 2013 Smith (SHSU) Elementary Functions 2013 1 / 35 Smith (SHSU) Elementary Functions 2013 2 / 35 Quadratic functions Quadratic functions The graph of a quadratic polynomial is a parabola. Here is the standard parabola y = x 2 . All of the graphs of quadratic functions can be created by transforming the parabola y = x 2 . Just as we have standard forms for the equations for lines (point-slope, slope-intercept, symmetric), we also have a standard form for a quadratic function. Every quadratic function can be put in standard form , f ( x ) = a ( x − h ) 2 + k where a, h and k are constants (real numbers.) Smith (SHSU) Elementary Functions 2013 3 / 35 Smith (SHSU) Elementary Functions 2013 4 / 35

Quadratic functions Quadratic functions The standard form for a quadratic polynomial is The turning point (0 , 0) on the parabola y = x 2 is called the vertex of the f ( x ) = a ( x − h ) 2 + k (1) parabola; it is moved by these transformations to the new vertex ( h, k ) . From our understanding of transformation, beginning with the simple y = x 2 , If we are given the equation f ( x ) = ax 2 + bx + c for a quadratic function, if we shift that parabola to the right h units, we can change the function into the above form by first factoring out the (replacing x by x − h ) term a so that f ( x ) = ax 2 + bx + c = a ( x 2 + b a x + c a ) . stretch it vertically by a factor of a (multiplying on the outside of the parentheses by a ) We then complete the square on the expression x 2 + b a x + c and then shift the parabola up k units, a (adding k on the outside of the parentheses) Let me explain completing the square.... we will have the graph for y = a ( x − h ) 2 + k. Smith (SHSU) Elementary Functions 2013 5 / 35 Smith (SHSU) Elementary Functions 2013 6 / 35 Completing the square Completing the square – the main idea Here is a test of our understanding of the simple equation A major tool for solving quadratic equations is to turn a quadratic into an ( x + A ) 2 = x 2 + 2 Ax + A 2 . Fill in the blanks: expression involving a sum of a square and a constant term. 1 ( x + 2) 2 = x 2 + 4 x + 4 This technique is called “completing the square.” 2 ( x + 5) 2 = x 2 + 10 x + 25 Recall that if we square the linear term x + A we get 3 ( x + 1) 2 = x 2 + 2 x + ( x + A ) 2 = x 2 + 2 Ax + A 2 . 1 − 1) 2 = x 2 − 2 x + 4 ( x + 1 Most of us are not surprised to see the x 2 or A 2 come up in the expression − 3) 2 = x 2 − 6 x + but the expression 2 Ax , the “cross-term” is also a critical part of our 5 ( x + 9 answer. 6 ( x + 7 49 2) 2 = x 2 + 7 x + In the expansion of ( x + A )( x + A ) we summed Ax twice. 4 Smith (SHSU) Elementary Functions 2013 7 / 35 Smith (SHSU) Elementary Functions 2013 8 / 35

Completing the square – the main idea Completing the square – the main idea Applying this idea. If one understands this simple idea, that we can predict the square by Since x 2 − 6 x is the beginning of looking at the coefficient of x , then we can rewrite any quadratic x 2 − 6 x + 9 = ( x − 3) 2 polynomial into an expression involving a perfect square. For example, since then x 2 + 4 x + 4 = ( x + 2) 2 x 2 − 6 x = ( x − 3) 2 − 9 then a polynomial that begins x 2 + 4 x must involve ( x + 2) 2 . and so x 2 − 6 x + 2 = ( x − 3) 2 − 7 , For example: x 2 + 4 x = ( x + 2) 2 − 4 x 2 − 6 x + 10 = ( x − 3) 2 + 1 , x 2 + 4 x + 7 = ( x + 2) 2 + 3 x 2 − 6 x − 3 = ( x − 3) 2 − 12 , x 2 + 4 x − 5 = ( x + 2) 2 − 9 Smith (SHSU) Elementary Functions 2013 9 / 35 Smith (SHSU) Elementary Functions 2013 10 / 35 Completing the square – the main idea Completing the square – the main idea This is useful if we are trying to put an equation y = f ( x ) of a quadratic Since x 2 − 3 x is the beginning of into the standard form y = a ( x − h ) 2 + k. x 2 − 3 x + 9 4 = ( x − 3 In the next presentation, we apply the concept of “completing the square” 2) 2 to put a quadratic into standard form and then we will use that form to develop the quadratic formula. then x 2 − 3 x = ( x − 3 2) 2 − 9 4 . (END) Smith (SHSU) Elementary Functions 2013 11 / 35 Smith (SHSU) Elementary Functions 2013 12 / 35

Completing the square – the main idea In the last lecture we discussed quadratic functions and introduced the concept of “completing the square”, rewriting x 2 + 4 x + 7 = ( x + 2) 2 + 3 Elementary Functions Part 2, Polynomials x 2 + 4 x − 5 = ( x + 2) 2 − 9 Lecture 2.1b, Applications of completing the square x 2 − 6 x + 9 = ( x − 3) 2 Dr. Ken W. Smith x 2 − 6 x = ( x − 3) 2 − 9 . Sam Houston State University 2013 x 2 − 6 x + 2 = ( x − 3) 2 − 7 , x 2 − 6 x + 10 = ( x − 3) 2 + 1 , x 2 − 6 x − 3 = ( x − 3) 2 − 12 . Smith (SHSU) Elementary Functions 2013 13 / 35 Smith (SHSU) Elementary Functions 2013 14 / 35 Completing the square – the main idea Completing the square – the main idea For example, if y = x 2 + 4 x + 7 Since x 2 − 3 x is the beginning of is the equation of a parabola then we can complete the square, writing x 2 + 4 x = ( x + 2) 2 − 4 and so x 2 + 4 x + 7 = ( x + 2) 2 − 4 + 7 = ( x + 2) 2 + 3 . Our equation is now x 2 − 3 x + 9 4 = ( x − 3 2) 2 y = ( x + 2) 2 + 3 . then x 2 − 3 x = ( x − 3 2) 2 − 9 4 . This is the standard form for the quadratic. We see that the vertex of the parabola is ( − 2 , 3) . We can transform the graph of y = x 2 into the graph of y = ( x + 2) 2 + 3 by shifting the graph of y = x 2 two units to the left and then shifting the graph up 3 units. Smith (SHSU) Elementary Functions 2013 15 / 35 Smith (SHSU) Elementary Functions 2013 16 / 35

Completing the square – the main idea Completing the square – the main idea An extra step occurs if the coefficient of x 2 is not 1. What is the vertex of the parabola with equation y = 2 x 2 + 8 x + 15 ? How do we complete the square on something like 2 x 2 + 8 x + 15 ? Once again we focus on the first two terms, 2 x 2 + 8 x. Factor out the coefficient of x 2 so that Solution. Complete the square to write y = 2 x 2 + 8 x + 15 = 2( x + 2) 2 + 7 . 2 x 2 + 8 x = 2( x 2 + 4 x ) . The parabola with equation y = 2( x + 2) 2 + 7 has vertex ( − 2 , 7) . Now complete the square inside the parenthesis so that x 2 + 4 x = ( x 2 + 4 x + 4) − 4 = ( x + 2) 2 − 4 . Therefore To transform the parabola y = x 2 to the graph of y = 2( x + 2) 2 + 7 2 x 2 + 8 x = 2( x 2 + 4 x ) = 2[( x + 2) 2 − 4] = 2( x + 2) 2 − 8 . 1 first shift y = x 2 left by 2 units 2 then stretch the graph vertically by a factor of 2 If y = 2 x 2 + 8 x + 15 then completing the square gives 3 and then slide the graph up 7 units. y = 2 x 2 + 8 x + 15 = 2( x + 2) 2 − 8 + 15 = 2( x + 2) 2 + 7 . Smith (SHSU) Elementary Functions 2013 17 / 35 Smith (SHSU) Elementary Functions 2013 18 / 35 Completing the square – the main idea Completing the square – the main idea Find the standard form for a parabola with vertex (2 , 1) passing through (4 , 5) . Find the vertex of the parabola with graph given by the equation y = 2 x 2 − 12 x + 4 . Solution. A parabola with vertex (2 , 1) passing through (4 , 5) has standard form y = a ( x − h ) 2 + k where ( h, k ) is the coordinate of the vertex. Here h = 2 and k = 1 and so we know y = a ( x − 2) 2 + 1 . Solution. Completing the square, we see that 2 x 2 − 12 x + 4 = 2( x 2 − 6 x ) + 4 = 2(( x − 3) 2 − 9) + 4 By plugging in the point (4 , 5) we see that 5 = a (4 − 2) 2 + 1 = 2( x − 3) 2 − 18 + 4 = 2( x − 3) 2 − 14 . and so a = 1 . So our answer is So the vertex is (3 , − 14) . y = 1 ( x − 2 ) 2 + 1 Smith (SHSU) Elementary Functions 2013 19 / 35 Smith (SHSU) Elementary Functions 2013 20 / 35