Once you calculate the X coordinate using the axis of symmetry (X=-b/2a), you plug that value in for all of the X's in the equation of the parabola. You then solve the equation for the value of Y.
a dot
X= -b / 2a
The axis of symmetry for a parabola of the form y = ax2 + bx + c is x = -b/2a So the axis is x = -2/2*(-3) or x=1/3
18
The axis of symmetry is x = -2.
axis of symmetry is x=0 Vertex is (0,0) So the answer is : YES
Vertex = (0,0) Line of symmetry = y axis You should of known that as this function is only X^2
In the form y = ax² + bx + c the axis of symmetry is given by the line x = -b/2a The axis of symmetry runs through the vertex, and the vertex is given by (-b/2a, -b²/4a + c). For y = 2x² + 4x - 10: → axis of symmetry is x = -4/(2×2) = -4/4 = -1 → vertex = (-1, -4²/(4×2) - 10) = (-1, -16/8 - 10) = (-1, -12)
It is (-1, 3).
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
2
By completing the square y = (x+3)2+1 Axis of symmetry and vertex: x = -3 and (-3, 1) Note that the parabola has no x intercepts because the discriminant is less than zero
How about y = (x - 2)2 = x2 - 4x + 4 ? That is the equation of a parabola whose axis of symmetry is the vertical line, x = 2. Its vertex is located at the point (2, 0).
Assume the expression is: y = x² - 6x + 5 Complete the squares to get: y = x² - 6x + 9 + 5 - 9 = (x - 3)² - 4 By the vertex form: y = a(x - h)² + k where x = h is the axis of symmetry x = 3 is the axis of symmetry.
D
Take the derivative: y' = 2x + 4. Set equal to zero: 2x + 4 = 0 --> x = -2. Substitute into original and y = -3. Axis of symmetry is x = -2. Vertex is (-2,-3)