Find a quadratic function y=ax^2+bx+c

Question

whose graph passes through the points (1,3),(3,-1) and (4,0).

Answer 1

The points are (1,3), (3,-1) and (4,0)

The quadratic equation y  = ax ² + bx + c .

The above function passes through the points (1,3), (3,-1) and (4,0).

If (x , y ) = (1, 3) then 3 =  a (1)² + b (1) + c  -------> 3 = a + b + c -----> (1)

If (x , y ) = (3, -1) then -1 = a (3)² + b (3) + c -------> -1 = 9a +3b + c ----> (2)

If (x , y ) = (4, 0) then 0 = a (4)² + b (4) + c -------> 0 = 16a + 4b + c ----> (3)

Solve (1) and (2) to eliminate c - variable and obtain two variable equation.

eq(1) - eq(2) ----> 3 -(-1)  = (a + b + c ) - (9a + 3b + c )

4 = -8a - 2b  ----> (4)

Solve (2) and (3) to eliminate c  - variable and obtain two variable equation.

eq(2) + eq(3) ----> -1 - 0  = (9a + 3b + c ) - (16a + 4b + c )

-1 = -7a - b  ----> (5)

Solve (4) and (5) to eliminate b - variable and obtain one variable equation.

eq(4) - 2eq(5) ----> 4 - (-2) = (-8a - 2b) - (-14a - 2b )

6 = 6a ----> a = 1

Substitute a = 1 in equation 5 : -1 = -7(1) - b ----> -1 = -7 - b ----> b = -6

Substitute a = 1 and b = -6 in equation (2)

-1 = 9(1) +3(-6) + c ----> -1 = 9 - 18 + c ----> c = 8

Substitute c = 8, a = 1 and b = -6 in quadratic equation.

Required quadratic function is y = x  ² - 6x  + 8.

Graph

Choose values for x and find the corresponding values for y.

x	y  = x 2 - 6x + 8	(x, y )
2	y  = (2) 2 - 6(2) + 8 = 0	(2,0)
5	y  = (5) 2 - 6(5) + 8 =  3	(5,3)

1.Draw a coordinate plane.

2.Plot the coordinate points.

3.Then sketch the graph, connecting the points with a smooth curve.