SMH Secondary Math Honors Unit 4 Graphing Quadratic Functions 4.0 Forms of Quadratic Functions Form: ( ) f = a + b + c, where a 0. There are no parentheses. f = 3 + 7 Eample: ( ) Form: f ( ) = a( p)( q), where a 0. Written as a multiplication problem. Also known as intercept form. f = 4 + 5 Eample: ( ) ( )( ) Form: f ( ) = a( h) + k, where a 0. only shows up once, as part of a perfect square. Eample: ( ) ( ) f = + 7 1 Conic Form of a parabola: 4pp(yy kk) = ( h) or 4pp( h) = (yy kk) Eamples: 4(yy ) = ( + 5) or 8( + 6) = (yy 1) Eamples: State whether each quadratic function is in standard, factored, or verte form. f = + + 4 a) f ( ) = ( + 3)( 5) b) ( ) ( 4) f = + 5 c) ( ) d) f ( ) = + 5 e) f ( ) = 3( ) f) f ( ) ( ) = + 1 3 g) f ( ) = ( + 5) h) f ( ) = 3 + 4 i) f ( ) = 5
Vocabulary: 4.1 Graphing Quadratic Functions: Verte and Ais of Symmetry : The shape of the graph of a quadratic function. : A line that cuts a parabola in half. If you were to fold a parabola along its ais of symmetry, the two sides would overlap. The equation of the ais of symmetry looks like = #. : The tip of the parabola the point at which it changes direction. If the parabola opens up ( a > 0 ), the verte is the lowest point on the graph, or the minimum point. If the parabola opens down ( a < 0 ), the verte is the highest point on the graph, or the maimum point. a > 0 a < 0 Finding the 1. Plug in 0 for.. Simplify. Don t forget order of operations. Verte Form of a Quadratic Function: y = a( h) + k ( hk, ) = h The sign of h is the opposite of the sign in the equation. h moves the graph of y = right and left in the opposite direction as the sign in the equation (but the same direction as the sign of h itself). The sign of k is the same as the sign in the equation. k moves the graph of y = up and down in the same direction as the sign in the equation. o For y = ( ) + 5, h and k 5.,5 and the ais of symmetry is =. The graph of y o For y ( ) = = The verte is ( ) = moved right and up 5. = + 3 7, h 3 and k 7. = 3. The graph of = = The verte is ( 3, 7) y = moved left 3 and down 7. and the ais of symmetry is
Opens up if a is positive. Opens down if a is negative. Eamples: For each function, do the following: 1) State the coordinates of the verte. ) State the direction of the opening, that is, whether the parabola opens up or down. 3) Find the y-intercept. 4) Draw a rough sketch of the graph. 5) Find the Domain and Range a) ( ) y = 3 + + 7 Direction: ais of symmetry: sketch graph: Domain: Range: f = 1 + 4 6 Direction: ais of symmetry: sketch graph: c) ( ) ( ) Domain: Range: Vertical Stretch: a changes how wide or narrow the graph is. o If a > 1, the graph is narrower than the graph of o If a < 1, the graph is wider than the graph of y =. y =. Figure out the eact shape of the graph by making an,y table. Always use the verte as one point. Then choose two -values on each side of the verte to plug into the equation to find the corresponding y-coordinates. A shortcut is to use counting patterns to graph the parabola. Start at the verte, then count: 1, a, 4a If a is negative, count down instead of up. 3, 9a, etc. o For y = ( 3) 4, a =. Start at the verte ( 3, 4 ), and count 1, ;, 8; 3, 18
Eamples: Fill in the requested information for each function. Then draw the graph. a) ( ) ( ) f = 1 4 a = h= k = Domain: Range: b) ff() = ( + ) 1 a = h= k = Domain: Range: Standard Form: f ( ) = a + b + c Just like with the other forms, the graph opens up if a is positive and opens down if a is negative. b o The -coordinate of the verte is. (The opposite of b divided by times a) a o To find the y-coordinate, plug the -coordinate into the original equation.
Eamples: Fill in the requested information for each function. Then draw the graph. f = 8+ 17 a) ( ) a = b= c= What is the maimum/minimum value? Domain: Range: b) y 4 = + a= b= c= What is the maimum/minimum value? Factored Form: ff() = aa( pp)(yy qq) Like other forms, a is the vertical stretch (pp, 0) and (qq, 0) are the -intercepts (zeroes). The -value of the verte is eactly half-way between them Evaluate the function at = pp+qq to find the y value of the verte.
Eamples: Fill in the requested information for each function. Then draw the graph. a) ff() = ( 5)( + 1) pp = qq = direction of opening: verte: Is it ma or min: Domain: Range: b) yy = 1 ( + 3)( 1) pp = qq = direction of opening: verte: Is it ma or min: Domain: Range:
4. Graphing using zeroes, solutions, roots, and -intercepts of a Function: The values of that make f ( ) or y equal zero. If the zeros are real, they tell you the places where the graph crosses the -ais, or the -intercepts of the graph. Other words for zeros: solutions to f ( ) = 0, roots, -intercepts. Finding zeros (-intercepts): 1. Change y or f ( ) to 0.. Solve for. If the equation is in factored form, solving for is easy just think What would have to be to make each set of parentheses equal to 0? If the equation is in standard form, solve by factoring or by using quadratic formula If the equation is in verte form, get the perfect square by itself, take the square root of both sides (don t forget the ±), then solve for. If your answers are imaginary (negative under the square root), the graph doesn t have -intercepts. Eamples: For each equation, find the zeros and state whether the graph opens up or down. Then match the equation to the correct graph. a) y = ( 1)( + 3) b) f ( ) = 1 ( + ) c) y 3( 3)( 1) f = = d) ( ) ( ) Graph 1: Graph : Graph 3: Graph 4:
Eamples: For each function, do the following: 1) State whether the parabola has a maimum or minimum. ) State whether the parabola opens up or down. 3) Find the -intercept(s). 4) Find the y-intercept. 5) Draw a rough sketch of the graph. a) f ( ) ( 4)( 1) = + Ma/Min: Direction: -intercept(s): sketch graph: Graphing from Factored Form: 1. Determine whether the parabola will open up or open down.. Find the zeros or -intercepts. Let f( ) = 0 and solve the equation. Mark the -intercepts on the graph. 3. Find the y-intercept but substituting in = 0. Mark this point on the graph. 4. Find the ais of symmetry and the verte. Use the method from 4.1 If there is only one zero, then the ais of symmetry will run vertically through that point and that -intercept will also be the verte. 5. Use the pattern from Section 4.1 to find other points on the graph now that you have the location of the verte. **Remember that a parabola is a smooth curve. Do not draw straight lines! Eamples: Fill in the requested information for each function. Then draw the graph. f = + a) ( ) ( 1)( 3) Zeros (-intercepts): Domain: Range:
f = + 4+ 3 b) ( ) Factored Form: f ( ) = Zeros (-intercepts): f = + 4 c) ( ) Factored Form: f ( ) = Zeros (-intercepts):
f = + 1 16 d) ( ) Factored Form: f ( ) = Zeros (-intercepts): e) ff() = 6 + 15 Factored Form: f ( ) = Zeros (-intercepts):
4.3 Writing equations from a graph or from set of information Eamples: Write a quadratic equation or function in Verte Form: f( ) = y= a ( h) + k If you know the verte of a parabola, ( hk, ), then you still need at least one other point on the parabola in order to write an equation. Use the verte form and fill in all the information you have. Then use the point on the parabola and substitute in for and y. Solve for a. Write your final equation a) (,1 ), passes through ( 4,13 ) b) ( 5,3), passes through ( 1, 9) Eamples: Write the equation of each parabola based on the information in the graph. Follow the steps outlined above. Leave the equations in Verte Form. a) y 10 y 10 5 5-10 -5 5 10-10 -5 5 10-5 -5-10 -10
Eamples: Write a quadratic equation or function in Factored Form: f( ) = y= a ( p) ( q) Use the factored form if you know the roots (a.k.a. solutions, -intercepts, zeros). You will still need to know at least one other point on the parabola in order to write an equation. Use the factored form and fill in all the information you have. Then use the point on the parabola and substitute in for and y. Solve for a. Write your final equation a) Roots: ( 3,0 ) &(,0), goes through (, 4) b) -intercept: (3, 0) & (, 0), goes through (0, 1) = = goes through ( 6, 10) c) Zeros: 1 & 3, d) Roots: = 7 & = 7, goes through ( 6, 9) e) Solutions: 8 i & 8 i, = = passes through (, 04)
Eamples: Write the equation of each parabola based on the information in the graph. Leave the equations in Factored Form. y 10 a) b) y 10 5 5 0-10 -5 5 10-5 -5-10 -10-5 0 5 10-10 Eamples: Write a quadratic equation or function in Standard Form: ff() = aa + bbbb + cc First write the equation in either Verte Form or Factored Form (whichever seems easier) Then use correct order of operations to multiply/distribute in order to get rid of parenthesis List the three terms in correct order:, then, then the constant term a) Use the graph just above to write the equation in Standard Form b) Write one of the verte form equations in Standard Form c) Write the equation with the following characteristics in Factored Form, Verte Form and Standard Form (yes, we ll need to complete the square to get it into verte form ) roots at ( 3, 0) and (5, 0) and goes through the point (, 15)
4.4 Solving Quadratic Inequalities & Systems of Equations by Graphing Eamples: Solve each inequality using the graph of ( ) f = + 3. Notice that each of these inequalities involves the value of + 3, which is represented by the y-coordinate of the graph. In each case, we are trying to figure out what -values (-coordinates) make the inequality true. When trying to find where + 3 > 0, we are trying to figure out what -coordinates have a y-coordinate that is bigger than zero in other words, where is the graph above the -ais? f = + 3 a) + 3> 0 b) + 3 0 ( ) c) + 3< 0 d) + 3 0 Solving a Quadratic Inequality Using the Graph: 1. Write the inequality in standard form. Replace the inequality sign with an equal sign and solve the equation a + b + c = 0 by factoring, completing the square, or using the quadratic formula. This gives you the -intercepts of the graph of y = a + b + c.. Graph y = a + b + c. The graph does not have to be very detailed. A rough sketch of a parabola opening in the correct direction with the correct -intercepts is all you need. 3. The solutions of The solutions of The solutions of The solutions of a + b + c > 0 are the -values for which the graph is above the -ais. a + b + c 0 are the -values for which the graph is on or above the -ais. a + b + c < 0 are the -values for which the graph is below the -ais. a + b + c 0 are the -values for which the graph is on or below the -ais. 4. If the inequality involves or, the -intercepts are included in the solution set (use brackets). If the inequality involves < or >, the -intercepts are not included in the solution set (use parentheses).
Eamples: Solve each quadratic inequality and write the solution set in interval notation. a) ( 3)( + 1) 0 b) ( )( ) 7 5 < 0 c) + 5> 0 d) 4 1 0 e) < f) 4 0 + 10 9
g) + 10 7 h) > + 6 Solving Systems of Equations by Graphing Solving a system of equations means finding the values of and y that make both equations true. The solutions are usually written as ordered pairs ( y, ). Solving by graphing: 1. Solve both equations for y.. Graph both equations using y = m + b, transformations, or, y tables. 3. The points where the two graphs intersect (cross) are the solutions. 4. Write the solutions as ordered pairs. 5. Eamples: Solve by graphing. + y = 6 a) y = + 4 b) 6 y = 4 6 3y = 3