Determine the highest point of the curve y + x^2 - 3x - 16 = 0
The curve defined by the equation y + x^2 - 3x - 16 = 0 is a parabola. It can be rewritten as y = -x^2 + 3x + 16. The parabola opens downwards.
At the highest point of the curve, the tangent to the curve is parallel to the x-axis. The slope of the tangent is given by the derivative of y = -x^2 + 3x + 16. dy/dx = -2x + 3
The slope of a line parallel to the x-axis is 0. Solving -2x + 3 = 0 gives x = 3/2. The corresponding value of y for x = 3/2 is y = -9/4 + 9/2 + 16 = 73/4
The highest point of the given curve is (3/2, 73/4)
y + x^2 - 3x - 16 = 0
The vertex form of a parabola's equation is generally expressed as :
y= a(x-h)2+k. The vertex is (h,k)
Bring 16 to the other side by subtracting and begin to perform CTS (Completing the Square).
Remove the negative by multiplying both sides by -1.
And, begin to perform CTS (Completing the Square).
Replace 0 with y.
As we have written it as vertex form, h=3/2 and k=73/4.
Our highest point in this case if the vertex (3/2, 73/4)
The maximum coordinates should be (1.5, 18.25), of which, the second coordinate may be harder to plot on the actual graph, so round it to 18.3 when plotting the curve.
Your first step is to separate the variables:
`y = -x^2 + 3x + 16`
To find the maximum or minimum of a function, you must find where the derivative changes signs (equals zero). To find the derivative, you must use the "power rule."
`y' = -2x + 3`
`0 = -2x + 3`
Solve for "x":
`2x = 3`
`x = 3/2`
Plug in this value to your original equation.
`y = -(3/2)^2 + 3(3/2) + 16`
`y = 73/4`
Your maximum should be at the point (3/2, 73/4).
The give equation can be written as
Y= 3*X - X * X +16
A quadratic equation will have maxi or minima where first derivation of Y = 0
or 3 - 2*X = 0
as the secend derivative of Y is always -ve , so we can only have maxima and this occurs at X=1.5
The value of Y at X=1.5 is = 3* 1.5 - 1.5*1.5 +16 = ( 3-1.5)*1.5 +16= 18.25
Maxima point is (1.5, 18.25)
First you must rewrite this equation into standard form.
y + x^2 - 3x - 16 = 0
y = -x^2 + 3x + 16
(I moved everything instead of y to avoid the step of dividing by -1)
You can tell by the equation that this is a parabola and that because there is a negative in front, it opens downwards. You can either graph it or use calculus and take the derivative.
y' = -2x +3
Another way you could do it is by changing the equation to vertex form and if you know your parabola is facing downwards then your highest point would be the vertex (h,k).
First solve for y to get y=-x^2+3x+16. This is an upside down parabola.
Then take the derivative of the equation:
y' = -2x +3
Set this equation to zero and solve for x:
Theoretically, since this is an upside down parabola with no restrictions, we already know that the number we found has to be the max, but in the case that it wasn't obvious, you would prove it the following way. Make a number line, and test two numbers surrounding the one found to determine if it is a max or min point on this graph. Lets take the numbers x = 1 and 2
We know that if we plug these numbers into the y' equation, we get a positive # from x=1, which means a positive slope, and a negative # from x=2, which is a negative slope. Therefore giving us the requirements of a max on a graph.