find the vertex and x and y intercepts of the parabola?

Question

f(x) = 1/4(x+3)^2-4

Assume y = x^2

How do you find out the vertex and the x and y intercepts.

Answer 1

You didn't need the hint to assume y=x^2 as it's already apparent in the equation.
y-intercept is the easiest, as you simply plug an x value of zero into the equation you have (and you should get -7/4. Properly you should write the point as (0,-1.75)).
x-intercepts: Remember these are where the y value is zero, so rewrite the equation with the whole function equalling zero:

0 = 1/4(x+3)^2-4
4 = 1/4(x+3)^2
16 = (x+3)^2 <---- you could take the square root of
__________________both sides here but don't forget the
__________________roots could be positive or negative
__________________i.e. +/-4 = x + 3
__________________ +/-4 - 3 = x
__________________ so x = either 4-3 or -4-3

or you could continue:
16 = x^2 + 6x + 9
0 = x^2 + 6x - 7 <----- you could now reach for the
__________________ quadratic formula or factorise
0 = (x+7)(x-1)
so either x+7 = 0 or x-1 = 0
so either x = -7 or x = 1

So the curve cuts the x axis at the points -7 and 1
x intercepts are (-7,0) and (1,0)

The vertex:
You could use the formula that the vertex is the point which has an x value of -b/2a but its much easier in this case to look at where halfway between the x intercepts is. As the curve is symmetrical, the highest or lowest point will always have an x value which is exactly halfway between the two x intercepts.
Halfway between -7 and 1 is -3 (check it on a graph if you want), so the x value of the vertex is -3. You now simply plug
-3 into the equation you have:

f(-3) = 1/4(-3+3)^2-4
= -4

So the vertex is the point (-3,-4)

Answer 2

No, I assume y = f(x)
Thus
y+4 = (1/4)(x+3)^2
or
4(y+4) = (x+3)^2

Compare that with the standard
4ay = x^2, which has vertex (0,0) and focal length a.

The vertex is where
y+4 = 0 and x+3 = 0
i.e. (-3, -4)

You haven't asked this, but notice that since 4a =4,
therefore a = 1,
and so the focus is 1 unit above the vertex,
i.e. at (-2, -4)

The y intercept is always found by putting x = 0, and I'm sure you can do this for yourself.

For the x intercept, y = 0, and so you must solve
16 = (x+3)^2
Obviously there are two solutions (not surprising since a vertical parabola usually cuts the x axis at two points, if at all)
which you get by considering that both 4 and -4 give 16 when squared. You don't need me any more, do you.

PS. I don't get "best answer" for this one do I! You're working it from the function point of view, not
4ay = x^2,
so I guess the third answer is more in line with your approach.

Answer 3

First, get the main suitable equation on your parabola. x intercepts are at (-2,0) and (8,0). So... y = (x+2)(x-8) y = x^2 - 6x - sixteen Sub the y intercept (0,-sixteen) into the equation to make certain if any amendments must be made... -sixteen = 0^2 - 0 - sixteen -sixteen = -sixteen The formula satisfies the coordinates of the y intercept, x^2 - 6x - sixteen is the main suitable equation of the parabola. Now, we ought to get it into vertex form... what equation of the type (x - h) will jointly as squared supply x^2 - 6x + some thing? (x-3)^2 = x^2 - 6x + 9 Our well-known equation is x^2 - 6x - sixteen Substituting (x-3)^2 = x^2 - 6x + 9 into our well-known equation... y = x^2 - 6x + 9 - 25 ----> (-19 = 9-25) y = (x-3)^2 - 25 The above equation is the parabola in vertex form. in spite of satisfies (x-3)^2 = 0 is the x coordinate of the vertex. So, the x coordinate is 3. -25 is the y coordinate of the vertex. So the vertex is at (3,-25)

Answer 4

NO! I'm done with my with math so I will never do algebra again! Sorry pal, but your on your own for this one!