Find the vertex and graph of y=-x^2+6x-5.5?

Answer 1

y = -x^2 + 6x - 5.5
dy/dx = -2x + 6
Insist dy/dx = 0 to find local maximum.
0 = -2x + 6
2x = 6
x = 3
y = -(3)^2 + 6(3) - 5.5
y = -9 + 18 - 5.5
y = 3.5

The vertex is at (3, -3.5).

If you have a polynomial equation of the form,
y = a x^n + b x^(n-1) + ...
Then the first derivative is
dy/dx = an x^(n-1) + b(n-1) x^(n-2) + ...
When dy/dx is zero, the graph of y(x) has either a maximum or a minimum. In general, dy/dx is the slope of the line tangent to the curve of y(x) at x.

The neat thing about using differential calculus to find maxima and minima is it works for just about all (differentiable) functions, including polynomials and trigonometric functions. Once you know this trick, you can nearly always find those local maxes and mins. Duke95's answer is a special case that can be derived from the more general differential calculus method.

Answer 2

This is a quadratic equation in the form y=ax^2+bx+c where a=-1, b=6, and c=-5.5. If you know calculus, setting dy/dx equal to zero is a GREAT way to find the vertex. If you don't know calculus, you can still find it. The vertex is the high or low point of the parabola, expressed as an ordered pair (x,y). To find x, we can use the formula x=-b/(2a) so x=-(6)/[2(-1)]=3. Then to find y, plug that value for x back into the original equation.

To graph the equation, you will need a few more points.

Answer 3

You do realize people can't post graphics here, right?