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1.2 Graphs Exam Review. Graph equations of: lines in the form y = mx + c ( A ), all forms ( M ) parabolas in the form y = x 2 + c or factorized form eq : y = (x-2)(x+3) ( A ) Interpret the features of a graph in context ( A )
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Graph equations of: • lines in the form y = mx + c (A), all forms (M) • parabolas in the form y = x2+ c or factorized form eq: y = (x-2)(x+3) (A) • Interpret the features of a graph in context (A) • write equations for lines from a graph (A) • solve simultaneous equations by plotting on a graph • write equations for quadratics from a graph with a vertical scale factor – k value (M) , for a combination of 2 different types of transformations (E)
Graphing equations of a line Draw the graphs of: a) y = - 2x – 6 b) y = 8
Graph the equation of a parabola y = (x – 4)(x + 2) y = -x2 + 4
Interpret features of graphs in Context Jim gets quotes from the Mighty Minibus Co and Various Vansfor a minibus.Jim can use the graph below to work out the cost of a trip for each rental company. a) How much does the Mighty Minibus Co charge per kilometre? $1.40/km b) Charges by Various Vans include a fixed fee for each trip. How much is this fixed fee? c) For which length of trip do both companies charge the same? d) Write the equation for the Various Vans graph. $30 50 km y = 0.8 x + 30
Graphing lines in all forms 2y – 3x = 18 y x Intercept (set y=0) x = - 6 y Intercept (set x=0) y = 9 x
Solve Simultaneous Equations Plot each line on the same axis and find the point of intersection y = -2x – 4 y = x + 8 Where the 2 lines intersect is the solution: (- 4, 4)
Mere buys a small tent for her little sister for Christmas. The top part is modelled by a parabola. There are three zips: BE, DE, and EF. The equation of the frame of the top part of the tent is where yis the height in cm and xis the distance from the centre line in cm. a) What is the height of the top of the tent? b) What is the length of the horizontal zip, EF? c) The length of AC is 32 cm. Find the length of zip BE. At x=0, y = 80 cm At y =0, x = 40 cm, EF = 40 cm At x = 16, y = 67.2 cm
Excellence b) Two poles, PR and QS help keep the tent upright. They are 25 cm from the centre-line of the tent. The bottom part of the tent is also modelled by a parabola. It cuts the vertical axis at y = –20. The length FG = 10 cm. Find the length of the pole PR. - 48.75 Subst x = -25 into To get y = 48.75 cm Subst (40, -10) to solve for k in y = kx2 – 20 (from part a) k = 1/160 40 -25 -16.09 Subst x=-25 into y = x2 – 20 160 y= -16.09 cm The length of the pole PR is 48.75 + 16.09 = 64.84 cm