Graphs of Quadratic Functions Day 1 Graph Quadratic

Graphs of Quadratic Functions Day 1

Graph Quadratic Functions: Standard Form: y = 2 ax + bx + c Shape: Parabola When in standard form, If a is positive, the parabola opens up y = ax 2+bx+c If a is negative, the parabola opens down y = -ax 2+bx+c

Will It Open Up or Down? • y = 4 x 2 + 7 x – 4 • Up • y = -6. 5 x 2 + 9 • Down • -2 x 2 + y = 8 x + 1 y = 2 x 2 +8 x + 1 y=- *must be in standard form • Down • y + ½x 2 = -3 x ½x 2 - • Up 3 x *must be in standard form

Parts of Parabolas: Vertex: • Highest or lowest point of the graph (the max or min of the function) Axis of • Lies on the axis of symmetry Axis of Symmetry: Symmetry Vertical line of symmetry that divides parabola into two parts that are mirror images. ALL parabolas are Symmetric! Calculate with formula Vertex x intercepts: x intercepts • Roots of equation (the x= solutions)

Graphing Quadratic Functions Steps (make sure to identify the a, b, and c values) 1. Find the equation of the axis of symmetry & draw the vertical line on the graph 2. Find the vertex coordinates & plot vertex on axis of symmetry (plug x= value of axis of sym. into function and evaluate for y) 3. Find and plot at least 2 more points and their symmetric points (mirror image the same distance across axis of symmetry) -Use x-intercepts if possible (the roots of the function. Set equation =0 and solve for x. The solutions are the x-intercepts) -Use the y-intercept (c value) if possible -Pick another ‘easy but logical’ point (select any logical x value, plug in and evaluate for y) 4. Sketch the curve and reflect it across the xis of symmetry

Graph y = 2 x 1. Find the axis of symmetry. y = ax 2+bx+c a=1, b=0, and c=0 x=0 is the axis of symmetry 2. Find the vertex. x=0 is also the x value of the vertex, now find the y value. If x = 0, plug in y = (0)2 y = 0 Vertex = (0, 0)

Graph y = 2 x continued 3. Graph 2 more points Try the y-intercept (c value) y = ax 2+bx+c a=1, b=0, and c=0 y-intercept (0, 0) is also the vertex, so we don’t need to plot it again Find 2 other points and their symmetric point Select any x value you want and plug into function to find y value (make easy choices, ie. whole numbers) Select x=1 and x=2 y = (1)2 y = 1 makes point (1, 1) y = (2)2 y = 4 makes point (2, 4)

Graph y = 2 x continued 4. Graph all points and mirror images to make symmetric parabola axis of symmetry x=0 Vertex (0, 0) (1, 1) and mirror image (-1, 1) (2, 4) and mirror image (-2, 4) Check: Opens up because a is positive

Graph y = 2 x - 2 x - 3 1. Find the axis of symmetry. y = ax 2+bx+c a=1, b=-2, and c=-3 x=1 is the axis of symmetry 2. Find the vertex. x=1 is also the x value of the vertex, now find the y value. If x = 1, plug in y = (1)2 - 2(1) – 3 y = -4 Vertex = (1, -4)

Graph y = 2 x - 2 x – 3 continued 3. Graph more points Try the y-intercept (c value) y = ax 2+bx+c a=1, b=-2, and c=-3 y-intercept (0, -3) Find x intercepts Set the function =0 and solve for x. The solutions are the x intercepts 0 = x 2 – 2 x – 3 factor to solve 0 = (x– 3)(x+1) (*also called writing in factored form or writing as a product of 2 linear factors) x = 3 x = -1 makes points (3, 0) and (-1, 0)

Graph y = 2 x - 2 x – 3 continued 4. Graph all points and mirror images to make symmetric parabola axis of symmetry x=1 Vertex (1, -4) (0, -3) and mirror image (2, -3) (3, 0) and mirror image (-1, 0) Check: Opens up because a is positive

Graph Quadratic Example Notes: