Graphing & Algebraic Work

Section 2.1
Question 30



Using Desmos, I found the graph of f(x) to be as shown below. I can clearly see the axis of symmetry: x = -5. The vertex is also visible:
(-5, -11). However, the x-intercepts are not visible on this graph. I guess an actual graphing calculator is needed to find specific numbers.



Let me show the work algebraically.
Completing the square within the parentheses looks like a good start.

f(x) = (x^2 + 10x + 25 ) + 14 - 25

f(x) = (x + 5)(x + 5) - 11

f(x) = (x + 5)^2 - 11 is the standard form.

From the standard form, I clearly see the vertex to be (-5, -11). I also see the axis of symmetry to be
x = -5.

I now need to find the x-intercepts.

Let f(x) = 0 and solve for x.

0 = (x + 5)^2 - 11

11 = (x + 5)^2

sqrt{11} = sqrt{(x + 5)^2}

-sqrt{11} = x + 5

-5 - sqrt{11} = x

sqrt{11} = x + 5

-5 + sqrt{11} = x

Our x-intercepts are as follows:

-5 - sqrt{11} at the point (-5 - sqrt{11}, 0).

5 + sqrt{11} at the point (5 - sqrt{11}, 0).

You say?

 

question ask you to graph first and from the graph identify vertex, asymptote, and x-intercepts
THEN check it algebraically

 

Isn't that what I did?

 

no, you did not read the graph first to identify vertex, asymptote, and x-intercepts
you immediately did calculations
for example, from the graph you can see that x-intercepts are is at x=-8 or 9 and x=-1 or -2, then you calculate to make sure which guess is correct

 

The Desmos graph does not show details. This is why I decided to confirm algebraically. I will do one more on my own using a free graphing calculator app. Now, time to sleep. My miserable, rainy weekend is over. Back to work tonight.

 

the Desmos graph does not show details, but you need to take a closer look and guess

 

Guess? Really? Is that what this exercise is all about? Can you respond to my thread in the Algebra forum here zero^(zero)?

 

yes, it's about developing ability to read the graph

Zero to zeroth power is often said to be "an indeterminate form", because it could have several different values.

Since x^0 is 1 for all numbers x other than 0, it would be logical to define that 0^0 = 1.

But we could also think of 0^0 having the value 0, because zero to any power (other than the zero power) is zero.

Also, the logarithm of 0^0 would be 0*infinity, which is in itself an indeterminate form. So laws of logarithms wouldn't work with it.

So because of these problems, zero to zeroth power is usually said to be indeterminate. However, if 0^0 power needs to be defined to have some value, 1 is the most logical definition for its value.

In algebra and combinatorics, the generally agreed upon value is 0^0 = 1, whereas in mathematical analysis, the expression is sometimes left undefined.

 

Can you copy and paste this reply to the Algebra thread zero^(zero)?