Three functions are given below: f(x), g(x), and h(x). Explain how to find the axis of symmetry for each function and rank the functions based on their axis of symmetry (from smallest to largest).
f(x) = −2(x − 4)2 + 2 g(x) = 5x2 − 10x + 7 h(x)=
@esamalaa
Dee: May I ask you to please find the formula for "axis of symmetry." Hint: It's part of the quadratic formula. If you don't already know this formula, why not do an Internet search for "axis of symmetry"? Once you've typed in that formula here, I'd be h appy to continue working with you towards finding a solution to this problem.
I saw you were typing a response, but then stopped. ??
x=-b/2a
Great. Now you're given 3 quadratics, A, B and C (C is just a graph). Can you re-write f(x) = −2(x − 4)2 + 2 so that it's in standard form, y = ax^2 + bx + c? can y ou determine a, b and c for g(x) = 5x2 − 10x + 7? Can you identify the coordinates of the vertex of the parabola in C?
thats my problem..
Try "can y ou determine a, b and c for g(x) = 5x2 − 10x + 7?" first. This equation is already in standard form, so you just need to pick out the coefficients which represent the values of a, b and c.
5x2 (a) 10x (b) 7 (c)?
can y ou determine a, b and c for g(x) = 5x2 − 10x + 7? Match up the coefficients as follows: 5x2 − 10x + 7 ax^2 + bx + c Comparing the coeff. of the x^2 term, a=5. What are b and c? Be careful with the signs.
-10 & 7
Right. So, if x=-b?(2a) is the equation for finding the x-coordinate of the vertex, and a= 5, b=-10 and c= 7, please identify the x -coord of the vertex for this particular situation.
x=100
But wouldn't -b/(2a) be -(-10)/(2*5)? = to what?
ohhhhh i was multiplying so 1?
Yes, as 10/10 = 1. So, the x-coordinate of the vertex of the 2nd parabola is x=1. Now look at the graph that represents parabola C. What is the x-coord of the vertex there?
wait so f(x) is 1 also ?
No, sorry. Please, let's take care of the third parabola first; please lok at the graph carefully, find the vertex, and determine what the x-coord. of the vertex is. It's not 1.
-3.5?
-2. my bad!
Think of the vertex as being teh very tip of your nose. Instead of seeing -3.5, I see the vertex x-ccord. as being x=-2. yes.
(thinking)
We're now done with B and C. What about A? A is not a quadratic in std. form. We must muliply, add and subtract to put it into standard form, ax^2 + bx + c.
I'd like to see where you're coming from in terms of understanding this. What would you do first to put f(x) = −2(x − 4)2 + 2 into standard form?
distribute?
Well, yes, but first, please notice that you have the square of a binomial in there. Looking carefullyl, I've seen that you have (x-4)^2. That ^2 represents an exponent of 2.
Please go ahead: expand (x-4)^2. We'll deal with the other terms later.
x^2-8 ?
did you apply the FOIL method of expanding the square of a binomial? Your result should have 3 terms; currently it has only 2. What went wrong?
Note that (x-4)^2 is represented by (x-4)(x-4). Which ever is easier for you, please multiply that or (x-4)^2 out.
(x-4)(x-4)=x^2-8x+16
That's good. I'd suggest y ou write that down as a reminder of how to do this sort of problem in the future.
okay!
f(x) = −2(x − 4)2 + 2 then becomes -2(x^2-8x+16)+2. Would you do the indicated multiplication and then simplify the result to obtain this equation in std. form, please.
-2x^2+16x-30
Great! {please review that process, as you're likely to have to do it again. Now, remembering that x=-b/(2a), please find the x-coord of the vertex for parabola A. x=?
-2(A) 16(B) -30(C)
Yes (although I'd much prefer you write them as a=-2, b=16, c=-30).
thus, x= ??
4?
Very good.
Now, please write your x-coordinates for A, B and C here. For A it's x=4; for B, it's x = ? and for C, it's x= ???. Just go back and fetch these values from our previous discussion.
b=1 c=-2
Good; now, please arrange these 3 x-coordinates in ASCENDING order.
c=-2 b=1 a=4
right?
Except for a little house-cleaning, that's a great answer. You might write, "The x-coodinates of the 3 vertices, arranged in ascending order, are -2, 1 and 4, corresponding to parabolas C, B and A."
so, you're done now, and done well.
Are u OK with this discussion? Did it answer all of your questions and concerns?
Surely enjoyed working with you! All the best to you. MM
help me with "Roy exclaims that his quadratic with a discriminant of −9 has no real solutions. Roy then puts down his pencil and refuses to do any more work. Create an equation with a negative discriminant. Then explain to Roy, in calm and complete sentences, how to find the solutions, even though they are not real." please?
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