identify the vertex of the parabola y=4x2-40x+96
several ways to do this.
quickest way:
the x of the vertex is -b/(2a)
= -(-40)/8 = 5
sub back into equation
y = 4(25) -40(5) + 96 = -4
vertex (5,-4)
calculus way:
dy/dx = 8x - 40 = 0
x = 5
proceed as above
completing the square:
y = 4x^2 - 40x + 96
= 4(x^2 - 10x + 25 - 25 + 96
= 4((x-5)^2 - 25) + 96
= 4(x-5)^2 - 100 + 96
=4(x-5)^2 - 4
To identify the vertex of the parabola given by the equation y = 4x^2 - 40x + 96, you can use the formula:
x = -b / (2a)
where "a" and "b" are coefficients in the equation y = ax^2 + bx + c.
In this case, a = 4 and b = -40. Substituting the values into the formula, we get:
x = -(-40) / (2 * 4)
x = 40 / 8
x = 5
To find the y-coordinate of the vertex, substitute this value of x back into the equation:
y = 4 * (5)^2 - 40 * 5 + 96
y = 4 * 25 - 200 + 96
y = 100 - 200 + 96
y = -4
Therefore, the vertex of the given parabola is (5, -4).