I have a question in Algebra 2?

Question

I don't understand the hyperbola and prabloa. So to find the vertex, on equation is -b over 2a but how do u find the x coordinate? Then how do you draw it?

Answer 1

First, let's start from the beginning. The hyperbola and parabola are conic sections - a curve that you get when slice a single plane through a cone (or two cones in the case of a hyperbola).

To graph a parabola, we need to know the direction of the curve, the vertex, any points of intersection with the x- and y-axis, and then take a few test points.

Let's use the equation, y= (x^2)-x-12.

1. Find the direction of the curve.  

We know that the curve faces upward since the coefficient of x^2  is positive.

2. Find the vertex.

As you said in your question, we can use the equation -b/2a to find the the x-coordinate of the vertex of the parabola. Then plug in the x-coordinate to find the y.

x=-b/2a

x=-(-1)/2(1)

x=1/2

y= [(1/2)^2]-(1/2)-12= (1/4)-(1/2)-12= -49/4 or -12.25

So, the vertex of our parabola is a (1/2,-49/4).

2. Find the x-intercepts.  

So now we use are skills solving quadratic equations. This time we are lucky since our equation can be factored.

(x^2)-x-12=0

(x-4)(x+3)=0

x=4 and x=-3

Now, we know that parabola cross the x-axis at (4,0) and (-3,0).

4. Find the y-intercept.

To find the y-intercept, solve the equation for x=0.

y=(x^2)-x-12

y=-12.

So, the parabola cross the y-axis at (0,-12).

4. If you feel that you need some more points to help you draw the curve, solve the equation using test points.

Hyperbola

A hyperbola is more complex since it is a two part conic section. To graph it, we need to know, the direction of the curves, the coordinates for the foci and the equation for the asymptotes.

For example, let's us the equation

(x^2)/4 - (y^2)/9 = 1

1. Find direction of the curves.

Since the (x^2)/(a^2) term is the positive term, we know that the curves with intersect the x-axis.

2. Find the foci.

The foci will be at (-a,0) and (a,0). To find a, we take the square root of the denominator of the positive term.

a^2=4

a=√4 =2

So the foci are at (-2,0) and (2,0).

3. Find the asymptotes.

For hyperbola that cross the x-axis, the equation for the asymptotes is

y= ± (b/a)x

Thus, the asymptotes are at y= (3/2)x and y= -(3/2)x.

4. If you need more point to make sure your curve shape is accurate, make sure to use numbers that will get you answers in all the quadrants.

I hope this helps. The links below have more examples. I would especially suggest the one on hyperbolas since it goes step by step through all the standard cases.

Answer 2

ok, given the parabola

y = a x^2 + b x + c

then the vertex is the point (h, k) such that

h = -b/(2a) .. . . this is the x coordinate

k = a h^2 + b h + c .. . .. .    or you can use the formula

k = (4ac - b^2)/4a

to sketch the graph, the parabola is curving upward from the vertex if a is positive,

the parabola is curving downward from the vertex is a is negative.