Question

Quadric functions practice

Answer 1

1. Find the vertex of the parabola defined by the equation y = 2x^2 - 8x + 6.

To find the vertex of a parabola in the form y = ax^2 + bx + c, use the formula x = -b/(2a) to find the x-coordinate of the vertex. Then, substitute this x-value back into the original equation to find the y-coordinate.

In this case, a = 2, b = -8, and c = 6.

x = -(-8)/(2*2) = 8/4 = 2

Now, substitute x = 2 back into the equation:

y = 2(2)^2 - 8(2) + 6
y = 2(4) - 16 + 6
y = 8 - 16 + 6
y = -2

Therefore, the vertex of the parabola is (2, -2).

2. Determine the axis of symmetry of the parabola y = -3x^2 + 6x - 1.

The axis of symmetry of a parabola is a vertical line that passes through the vertex. The equation of the axis of symmetry is given by x = -b/(2a).

In this case, a = -3 and b = 6.

x = -6/(2*-3) = -6/-6 = 1

Therefore, the axis of symmetry of the parabola is x = 1.

3. Find the y-intercept of the parabola defined by the equation y = x^2 - 4x + 3.

To find the y-intercept, set x = 0 in the equation and solve for y.

y = 0^2 - 4(0) + 3
y = 0 - 0 + 3
y = 3

Therefore, the y-intercept of the parabola is (0, 3).