Find each of the following

F

frctl

Full Member
Joined
Jun 29, 2019
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252
(A)Intercepts
(B)Vertex
(C)Maximum or minimum
(D)Range

I am given:

Screen Shot 2020-03-03 at 10.02.01 AM.png

I think I need to convert this Point slope form to Standard form

How am I able to do this?
 
This is not in point-slope form; that applies to a line only. It is what some would call standard form for a quadratic function, and others would call vertex form.

But in this form, it is easy to see the vertex, the minimum or maximum, and the range; and it is not hard to find the intercepts. So don't bother converting (which you would do by distributing, sometimes called "FOIL"). Just use the form you were given.

Let's give it a start. What does it mean to ask for the y-intercept? What do you have to do to any function f(x) to find it?
 
In order to find the y-intercept, x must equal 0. Therefore applied to the given function f

f(0) = -(0 - 3)2+ 2
f(0) = 9 + 2
f(0) = 11

Alternatively I learned that the x-intercept = -b/2a
 
You have the correct y-intercept, which is, as you say, f(0).

But it is not the x-intercept that is -b/2a; that is the location of the vertex when the function is given in the form ax^2 + bx + c. (Not for the form you have.)

The x-intercept is the value of y for which f(x) = 0. Sound familiar?

I mentioned that the form you were given is sometimes called vertex form; that is because you can read the vertex directly from the equation. Does that sound at all familiar? What value of x would make f(x) as large or small as possible? (Hint: x^2 alone is lowest at x=0, because on either side of 0, x^2 is higher.)
 
This answer for the x-intercept appears incorrect since my answer key states x-int of 3 ± √2 and y-int of -7.

Finding the y intercept:
0 = -(x - 3)2 + 2
-2 = -x2 + 9
-11 = -x2
x = √11
 
You appear to have assumed that (x-3)^2 = x^2 + 3^2. That is not true.

The second line should just be -2 = -(x-3)^2

Now divide both sides by -1 to isolate the square, then take the square root of each side to eliminate the square.

I just realize I missed a mistake in your calculation of the y-intercept. You did the right thing, but you missed a negative. The second line in post #3 should have had -9, not 9.
 
Thank you, is it like follows
0 = -(x - 3)2 + 2
-2 = -(x-3)2
2 = (x-3)2
√2 = √(x-3)
2 = x-3
x = 5
 
No, the fourth line should be ±√2 = (x-3) . You took the square root twice, and forgot the plus or minus.
 
Can I simplify ±√2 = (x-3) to get 3 = ±√2? This means I have found the (A) Intercepts, now I must find (B), (C) and (D)
 
frctl said:
Can I simplify ±√2 = (x-3) to get 3 = ±√2? This means I have found the (A) Intercepts, now I must find (B), (C) and (D)
Click to expand...
How did you get that?

What is/are the value/s of x-intercept/s that you found?

Please be explicit.
 
I prefer to think along these line to solve (x-3)2=2. I asked myself what number or numbers, if any, can I square to get 2. The answer is \(\displaystyle
\frac{+}{-}\sqrt{2}\).

What what we are squaring, namely x-3 can equal one of those two number.

Now we solve \(\displaystyle x-3 = \sqrt{2} \ and \ x-3 = -\sqrt{2} \). Is that clear?
 
frctl said:
Can I simplify ±√2 = (x-3) to get 3 = ±√2? This means I have found the (A) Intercepts, now I must find (B), (C) and (D)
Click to expand...
No, simply because 3 = ±√2 is not true. Please read what you write. You wrote 3 equal plus/minus square root of 2. Do you believe that? Even if you do (and I am sure that you don't) what does x equal? You have No x?
 
Yes, you're right. The x intercept is therefore
x - 3 = √2
x = √2 + 3
and
x - 3 = -√2
x = -√2 + 3,
and the y intercept is -7.
 
To find the vertex of an equation in the form f(x) = a(x-b)2 + c you ask yourself two questions.

1) What makes x-b equal to 0 (ie solve x-b=0. So the answer is x=b

2) What is the constant? In our case it is c

So the vertex is at ( b, c).

Now you try this with your function.
 
f.png
is the vertex (3, 2)

The constant c must be 2

I still require the max/min and range
 
frctl said:
therefore max is the vertex

min is negative infinity
Click to expand...
Well, to be precise, we would say that there is no miminum for this function. (Infinity isn't a number.)

Otherwise well done!

-Dan
 
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