Graphs

Parabola

y = (x+3)2 y = x2 y = x2+1 y = x2-2 y = x2+3

Find the equation of the graph using The equation is y = a (x – 2 ) ( x + 5) Using the point ( 0 , -5) The final equation is y = ½ ( x – 2 ) ( x + 5 ) -5 2 -5 Find the equation of the graph using y = a ( x + b ) ( x + c )

y = ½ ( x – 2 ) ( x + 5 ) y = ½ ( x – 3 ) ( x + 4 ) y = ½ ( x – 2 ) ( x + 5 ) + 5

Hyperbola

y = or xy = 1 (1 , 1) Asymptote at y = 0 (-1 , -1)

Asymptote down x = 0 (1 , 3) (graph has Asymptote along y = 2 (-1 , 1) y=1/x Asymptote down x = 0 (1 , 3) (graph has moved up 2) Asymptote along y = 2 (-1 , 1)

Asymptote down x = 3 (-2 , 1) (graph has moved Asymptote along y = 0 y=1/x Asymptote down x = 3 (-2 , 1) (graph has moved to the left 3 places) Asymptote along y = 0 (- 4 , -1)

Exponential

y = 2x y = 3x (0,1) Exponential functions illustrate growth and decline. When x = 0, y = a0 = 1, so the y-intercept of basic exponential functions is (0,1) for any value of a. The graph has no x-intercept because the equation ax =0 has no solution.

y = 2x + 1 (0,2)

y = 2(x+1) y = 2(x+2)

Y-intercept is (0,2), so y = 2ax (1,6) is on the graph 6 = 2 x a1 a = 3 Equation: y = 2(3x)

Write the equations for the following graphs Y = 2x, exponential Some coordinates other than (0,1) ie.(1,2) (2,4) (3,8) General graph is y = ax putting (3,8) in we get 8 = a3 and solve for a. a=2 Then check with another point Write the equations for the following graphs y = (x+1)(x-2)2 One of the turning points is on the x axis and the other cutting through the x axis xy= - 4, hyperbola Drawn in second and forth quadrants therefore in the form xy = - a. Passes through points (-1,4) & (-2, 2)

Trig graphs

y = sin x y = sin ( x + π/2 )

y = sin ( x ) + 2

y = sin ( 2x )

y = 2 sin ( x )

y = cos x