1.) y=3x^2+1, 2.) y=x^2-18+12, 3.) y=3x^2+6x-8

Question

find the equation of the axis of symmetry and the coordinates of the vertex of the graph of each function.

Answer 1

1) given y = 3x^2+1

Compare it to standard form of parabola is y = ax^2+bx+c

a = 3, b = 0, c = 1

Axis of symmetry x = -b/2a = 0

Substitute the x value in given equation.

y = 1

Vertex is (0,1)

Graph

Answer 2

2) given y = x^2-18+12

y = x^2-6

Compare it to standard form of parabola is y = ax^2+bx+c

a = 1, b = 0, c = 6.

Axis of symmetry x = -b/2a = 0

Substitute the x value in given equation.

y = -6

y = -6

Vertex is (0,-6).

 

Comments

plz give me answer for third question

Answer 3

3) y  = 3x ² + 6x - 8

Compare it to standard form of parabola equation y = ax ² + bx + c

Axis of symmetry x = -b /2a  = -6/6 = -1.

To find vertex of parabola substitute x  value in y = 3x ² + 6x - 8

y  = 3(-1)^2 + 6(-1) -8 = -11

Vertex = (-1, -11)

Make the table of values to find ordered pairs that satisfy the equation.

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

x	y = 3x 2 + 6x - 8	(x, y )
1	y =  3(1) 2 + 6(1) - 8 = 1	(1,1)
-1	y = 3(-1) 2 + 6(-1) - 8 = -11	(-1,-11)
-2	y = 3(-2) 2 + 6(-2) - 8 = -8	(1,-2)
0	y = 3(0) 2 + 6(0) - 8 = -8	(0,-8)
-3	y = 3(-3) 2 + 6(-3) - 8 = 1	(-3,1)

Graph

1.Draw a coordinate plane.

2.Plot the coordinate points found in table and axis of symmetry .

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

Axis of symmetry x = -1

Vertex of parabola (-1, -11).