can someone help me?

Question

without drawing a graph of the given equation,can you determine,
(a) how many x intercepts the parabola has
(b) whether its vertex lies above or below or on the x-axis.

Answer 1

Let y=f(x) be the function in question in an Oxy coordinated system.

(a) The parabola has either none, one or two intercepts. Solving f(x)=0 will determine the exact intercepts, since any point on the x-axis has y = f(x) = 0. If f(x) = ax^2 + bx + c, calculate D=(b^2-4ac). If:

1.D<0: f(x) does not intercept the x-axis.
2. D = 0: f(x) has only one intercept.
3. D>0: f(x) has two intercepts.

(b) If you are familiar with derivatives, the problem is easy. At the vertex, the slope of the tangent to f(x) is zero. Therefore:

f ' (x) = 2ax + b = 0, and x = -b/2a. This is the x coordinate of the vertex. Determine x = -b/2a and then determine the y-coordinate by replacing x into y=f(x). Thereafter:

1. If y=0: vertex is on the x-axis
2. If y>0: vertex is above the x-axis
3. If y<0: vertex is below the x-axis.

You can also check your calculations if the vertex is on the x-axis: if that is the case, the vertex IS the intercept of f(x) with the x-axis. Be careful: if y>0, this means the vertex is above the x-axis, however there may be two or no intercepts on the x-axis, depending on whether the "bowl" of f(x) is facing down or up.

Hope this helps.

Answer 2

The parabola can have 1) One intercept if y=x^2 (x=0)
(2) Two intercepts otherwise (the zeros of the equation). The reason you have two is that a parabola is a quadratic function.

For the second part, if the equation is of the form y = x^2 + a, the graph will be lie above the x axis if a is a positive number; it will lie below the x axis is a is a negative number.

You may want to try the case x = y^2 on your own.

Answer 3

Set the y value equal to zero. If you know the quadratic formula, find out how many real solutions there are to the equation. The number of real solutions is the same as the number of x intercepts.

If the x^2 part is positive and there are two real solutions, the vertex must be below the x axis. If there is one real solution, then the vertex is on the x axis. If the x^2 part is positive and there are no real solutions, then the vertex is above the x axis.

If the x^2 part is negative, and there are two real solutions, the vertex is above the x axis. If the x^2 part is negative, and there are no real solutions, then the vertex is below the x axis.

Answer 4

Usually these sort of questions are asked about parabolas that are either opening upward or downward, which is to say that y = something x squared Plus or minus something x plus or minus something (usually written as y = ax^2 + bx + c). If that's the case, you can tell both things by the quadratic formula x = -b +- sqrt (b^2-4ac) all over 2a.

If b^2 - 4ac > 0, there are 2 x-intercepts
If b^2-4ac = 0 there is 1.
If b^2-4ac < 0 there are none.

The vertex occurs when x = -b/(2a) which is the Q.F. without the +- part. Plug this value in for x to get y, and if it's positive the vertex is above the x-axis; negative, below; zero, on.

Answer 5

sorry nope i ant even in high school

Answer 6

don't know