You can work this out by taking the derivative of the equation, and solving for zero:
y = x2 - 8x + 18
y' = 2x - 8
0 = 2x - 8
x = 4
So the vertex occurs where x is equal to 4. You can then plug that back into the original equation to get the y-coordinate:
y = 42 - 8(4) + 18
y = 16 - 32 + 18
y = 2
So the vertex of the parabola occurs at the point (4, 2), leaving 2 as the answer to your question.
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).
The given equation is not that of a parabola since there are no powers of 2. Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals" etc. And using ^ to indicate powers (eg x-squared = x^2).
7
Since no points were given, for any point (x,y), plug the x and y values of the point into the equation. If you get a contradiction, ie 5=3 or something similar, then the point does not lie on the graph.
7
The equation does not represent that of a parabola.
It is the equation of a parabola.
It is the parabola such that the coordinates of each point on it satisfies the given equation.
No you can't. There is no unique solution for 'x' and 'y'. The equation describes a parabola, and every point on the parabola satisfies the equation.
1/([*sqrt(cx)]
X equals 0.5at squared is a quadratic equation. It describes a parabola. Y equals mx plus b is a linear equation. It describes a line. You cannot describe a parabola with a linear equation.
An equation of a parabola in the x-y plane, is one possibility.
At: x = 6
When x = -5
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).
I'm guessing you mean y = x², which is the equation of a parabola. There is no one answer. Every point which lies on the parabola is the solution set to the equation. Some examples of points which satisfy this are: (0,0) (1,1) (2,4) (-3,9) (-5,25) and (½,¼)
The equation y = 4x^2 + 5 is a parabola