Expert: Paul Klarreich - 4/25/2007

Question
Hiya i wanted to sketch a graph of the parabolic curve, but before i wanted to
solve the quadractic equation first.

i need to see how your answer ends with these....

please advice

question1

solve the quadractic equation, in each case sketch a graph of the parabolic
curve.

(i) 4x^2 -144=0

(ii)  x^2-20x+91=0

(ii) x^2-4x-21=0

For the quadratic equation, i wanted to know the final equation to use to draw
table and graph.



Question 2

Sketch graphs of the function.

i wanted to see you final equation for these questions.i know how to work it
out i just desperately need to know its correct.

(i) (x+2)(x+1)(x-4)

(ii) (x-2)(x+4)(x-6)

(iii) y=e^2x


hope to hear from you.

regards

pearl


Answer
Questioner:   pearl Addo
Category:  Advanced Math
Private:  No
 
Subject:  sketching graphs with equations
Question:  Hiya i wanted to sketch a graph of the parabolic curve, but before i wanted to
solve the quadractic equation first.

i need to see how your answer ends with these....

please advice

question1

solve the quadratic equation, in each case sketch a graph of the parabolic
curve.

(i) 4x^2 -144=0

(ii)  x^2-20x+91=0

(ii) x^2-4x-21=0

For the quadratic equation, i wanted to know the final equation to use to draw
table and graph.



Question 2

Sketch graphs of the function.

i wanted to see you final equation for these questions.i know how to work it
out i just desperately need to know its correct.

(i) (x+2)(x+1)(x-4)

(ii) (x-2)(x+4)(x-6)

(iii) y=e^2x
hope to hear from you.
regards
pearl
...............................................
Hi, Pearl,

I ask questioners to tell me what math they are studying because in questions like these, I might use some methods that you are not studying, and that doesn't do you any good.  I will assume you are studying precalculus.

For these three:

(i) 4x^2 -144=0

(ii)  x^2-20x+91=0

(iii) x^2-4x-21=0

Unfortunately, I can't make a graph for you, but I can indicate some of its properties and leave the sketching to you.

Keep in mind that:

4x^2 - 144 = 0  is a quadratic equation, while
y = 4x^2 - 144  is a quadratic function, which is a different thing.

Solving the equation can help, of course -- it gives you the x-intercepts.  Once you have those, if you take their average (remember how to do averages?  add and divide by 2) you get the x-coordinate of the vertex.

(i)  4x^2 - 144 = 0
   4(x^2 - 36) = 0
4(x - 6)(x + 6) = 0
x = +- 6  are the solutions and thus the x-intercepts for the graph.

In y = 4x^2 - 144, the  x^2 term is positive, so the parabola will open upward.  
When x - 0, y = -144, so that is the  y-intercept.  

Mark the points  (+-6, 0) and (0,-144).  [Yes, minus 144]
Draw a parabola that is opening upward and through those points.

.............................

(ii)  x^2-20x+91=0
Factor:

(x - 13)(x - 7) = 0
x = 7, x = 13  are the x-intercepts.
The vertex is at  x = 10, the average of 7 and 13.  Take:

y = x^2 - 20x + 91,
substitute  x = 10:
y = 100 - 200 + 91 = -9

Mark the intercepts and vertex, etc.

(iii) x^2-4x-21=0

I think this works out the same way, so I'll leave it to you.

.....................................................

For your part 2:

(i) (x+2)(x+1)(x-4)  is nothing at all.  Perhaps you meant to write:

y = (x+2)(x+1)(x-4)
The intercepts here are  x = -2, x = -1, and x = 4.
This is a cubic.  The x^3 term will be positive, so it starts way down in quadrant 3, and ends in quadrant 1.  You can handle the rest.

(ii) (x-2)(x+4)(x-6)

Same stuff here.

(iii) y=e^2x

This is an exponential function.  It will:

A. always be positive, so the graph will always be above the x-axis.
B. always be rising.
C. be asymptotic to the negative x-axis on the left side.
D. get real high real fast on the right side.
E. pass through  (0,1)

I think that should do it.