Find the coordinates of the vertex and the equation of the axis of symmetry of the parabola with the given equation.
y = 8x^2 + 48x − 3
vertex
(x, y) =
axis of symmetry ________
y = 8(x^2+6x) - 3
now just complete the square.
y = 8(x^2+6x+9) - 3
but wait. You just added 8*9, so subtract it off so things don't get changed.
y = 8(x^2+6x+9) - 3 - 8*9
y = 8(x+3)^2 - 75
now it's easy to see that the vertex is at (-3,-75), and of course, it lies on the axis of symmetry.
To find the coordinates of the vertex of a parabola, we can use the formula -b/2a, where a and b are the coefficients of the quadratic equation in the standard form ax^2 + bx + c = 0. In this case, the equation of the parabola is y = 8x^2 + 48x − 3, so the coefficients a = 8 and b = 48.
First, let's find the x-coordinate of the vertex using the formula -b/2a:
x = -48 / (2 * 8) = -3/2
To find the y-coordinate of the vertex, substitute the x-coordinate (-3/2) into the original equation:
y = 8(-3/2)^2 + 48(-3/2) − 3
y = 8(9/4) - 72/2 - 3
y = 18 - 36 - 3
y = -21
So, the coordinates of the vertex are (x, y) = (-3/2, -21).
To find the equation of the axis of symmetry, it is a vertical line that passes through the vertex. The equation of a vertical line is given by x = a, where 'a' is the x-coordinate of the vertex.
Therefore, the equation of the axis of symmetry for this parabola is x = -3/2.