Systems of Linear and Quadratic Equations

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Presentation transcript:

Systems of Linear and Quadratic Equations Find their intersections.

What does it mean to solve a system? To solve a system, we find where the graphs intersect. You know how to solve a system that contains linear equations. The two lines intersect at ONE point.

How about the intersection of a line with a quadratic equation? The graph of a quadratic equation is a parabola.

How many points of intersection does a line have with a parabola? Could have No Solutions Could have One Solution Could have Two Solutions

To solve the system graphically: Analyze the graph. Determine where the line intersects the parabola. Write down the ordered pairs. Solution: (1, 3) and (6, 13)

Solve these systems: (-3, 4) and (1, 0) (-2, -2) and about (2.5, 2.5)

How to solve the system algebraically: Make both equations into “y =“ format. Set them equal to each other. Solve for x. You should get 2 values for x. Substitute both values back into EITHER equation and solve for y. Write your answers as ordered pairs.

Solve this system: y = x2 + 2x and y = 2x + 4 Substitute each x- value into either equation and solve for y. y = 2(2) + 4 y = 8 y = 2(-2) + 4 y = 0 Since both equations are already “y =“, set them equal to each other. x2 + 2x = 2x + 4 -2x -2x x2 = 4 Square root both sides. x = 2 and -2 Solution: (2, 8) and (-2, 0)

Solve this system: y = 3x2 + 4(x-1) and y = 4x +23 Since both equations are already “y =“, set them equal to each other. 3x2 + 4(x-1) = 4x + 23 3x2 + 4x - 4 = 4x + 23 -4x -4x 3x2 – 4 = 23 3x2 = 27 x2 = 9 Square root both sides. x = 3 and -3 Substitute each x- value into either equation and solve for y. y = 4(3) + 23 y = 35 y = 4(-3) + 23 y = 11 Solution: (3, 35) and (-3, 11)