I Need MAJOR HELP i dont know how to do any of these our long term sub can not explain to me in away i understand please help!

Question

Okay so my sheet say,

"write the equation for each parabola in standard form and finf the vertex of the parabola"

 

My Equations are like this:

 

1.  y=(x+4)(x-2)

 

Answer 1

Dear Mary;

 

In general, the standard form of a parabolic equation is y = ax² + bx + c and the vertex form of the same equation is y a(x - h)² + k

 

 

1. y =(x+4)(x-2) can be rewritten as y = x² + 2x - 8 which is the "standard form."

 

The formula for finding the x-coordinate of the vertex is (-b/2a). In this case (-2/(2)(1) = -1.   So, the axis of symmetry for our parabola is x = -1.  To find the y-coordinate of the vertex, merely substitute x = -1 back into the equation of the x² + 2x - 8 which gives (-1)2 + (2)(-1) - 8 or -9.

 

So the vertex of this parabola is (-1, -9)

 

3. 3(x+3)(x+5) = 3x² + 24x + 45

 

so a = 3, b = 24 and c = 45

 

x-coordinate of vertex = (-b/2a) = (-24)/[(2)(3)] = -4

 

y-coordinate of vertex = (3)(-4)² + (24)(-4) + 45 = 48 - 96 + 45 = -3

 

Vertex is at (-3, -4)

 

2. y = -(x+9)² + 41 = - x² - 18x -81 + 41 = -x² - 18x - 40

 

a = -1, b = -18, c = -40

 

x coordinate of vertex = -b/2a = -(-18)/[(2)(-1)] = -9

 

y coordinate of vertex = -(-9)² - (18)(-9) - 40 = 41

 

Vertex = (-9, 41)

 

Mary, now that I've gotten you started, do you feel confident enough to do the rest on your own?  If not, contact me again.  I wish I could be the "long term substitute" for your class!

 

Here is a very useful website for you to visit: http://www.endmemo.com/geometry/parabola.php

Answer 2

The standard form of a parabola is  

 

y =   a x² + b x  + c             where the parameters a,b,c are numbers that do not involve x

 

The first step is reorganize the given equation into this form.    I will  examine the first equation to show how this is done.  We start with  

 

y = (x+4)(x-2)         the FOIL rule (first, outer,inner,last) or repeated use of the distributive law gives

 

y =  x² -2x + 4x -8           collecting like terms  gives

 

y =  x² + 2x   - 8         this is now in the standard form with   a = 1, b = 2.  c = -8

 

The  x and y  coordinates of the vertex of a parabola in standard form are

 

vertex   =  (  -b/(2a) ,   -b²/(4a) +c  )   since, for the first equation,  a=1, b=2, c= -8    we have

 

vertex = ( -1 , -9)

 

Notice that if you substitute x = -1 into the original equation you get y = -9.    This is the lowest point on this upward opening parabolic curve.    More generally, the vertex is the lowest point on an upward opening parabola and, for a downward opening parabola, the vertex is the highest point on the curve.

 

 

The rest of the  equations can be worked using the same procedure.

 

 

 

Answer 3

The standard form equation of a parabola is a quadratic equation in the form:

f(x) = a x^2 + b x + c.

 

The simplest quadratic equation is f(x) = x^2 (opens up, vertex at origin). Axis of Symmetry is x = 0 (the y-axis).

 

If you scale with a factor of a, you get f(x) = a x^2. Thinner if |a| > 1, fatter if |a| < 1, opens up if a > 0, down if a < 0. Vertex at origin. Axis of Symmetry is x = 0 (the y-axis).

 

If you translate f(x) = a x^2 by the translation vector <h,k> you get the Vertex Form:

f(x) - k = a (x - h)^2, or f(x) = a (x - h)^2 + k. The vertex is at (h,k). Axis of Symmetry is x = h (a vertical line).

 

If you multiply out the Vertex Form:

f(x) = a x^2 - 2ah x + ah^2 + k

and compare to standard form:

f(x) = a x^2 + b x + c

you can see that b = -2ah, or h = -b/(2a).

 

So if you have the standard form and want the vertex, (h,k), you can find h = -b/(2a), then find k = f(h). I.e., substitute h into f(x) to get k.

 

A few words on the factored form: f(x) = a (x - z_1) (x - z_2)

 

f(x) = 0 if x = either z_1 or z_2. If real the z's are x-intercepts; they won't be real if parabola doesn't intersect the x-axis.

 

The axis of symmetry is at x = (z_1 + z_2)/2 = h. Then k = f(h).

 

Let's do a couple of your problems now:

 

10. y=-1(x+4)(x-4)-8

 

Have to multiply out to get standard form first:

y = -1(x^2 - 16) - 8

y = - x^2 + 8

We see that b = 0 so h = 0 and k = y(0) = 8, vertex being (0,8);

or, the equation is easily converted to Vertex Form by replacing x with x - 0:

y = - (x - 0)^2 + 8

h = 0, k = 8, so vertex is (0,8).

 

5. y = -(x+9)^2+41

In Vertex Form, so vertex is (-9,41). Notice it's negative 9.

Just have to multiply it out to get standard form:

y = -(x^2 + 18 x + 81) + 41

y = - x^2 - 18 x - 81 + 41

y = - x^2 - 18 x - 40