Respuesta :
Alright. So this question looks super intimidating at first, but for questions like this, it's super important to break it down into steps. So, let's just focus on the first part. It wants us to transform y= 2x^2 - 12x + 23 into something that looks like y = 2 (x - c) ^2 + d
It's understood that c and d just stand for constants, or regular numbers without the variable x attached. So! How do we do that? First off, we take out a 2 from JUST 2x^2 and -12x. We can write this as
y = 2 (x^2 -6x) + 23
At this point, we need to recall that we want that (x^2 - 6x) part to be a "perfect square" or that when we factor it, the two factors are the same thing. So, if we wanted x^2 - 6x to be a perfect square, what would we theoretically need to add onto it? To further explain this, I'm gonna rewrite the above quadratic equation as this:
x^2 - 6x + w
Where w is the unknown value that will make this quadratic a perfect square. So, what would w have to be to make this a perfect square? well, if this were just a normal quadratic, we'd need to find two numbers that multiply together to give me the coefficient of x^2 × w, which is, in this case, 1×w, or just w. Those same two numbers need to add together to give me -6.
( ) ( ) = w
( ) + ( ) = -6
We also know these two numbers have to be the SAME, so we factor into a perfect square. The only number, when added to itself, that equals -6, is -3. So, our two mystery numbers have to be -3 and -3.
(-3)(-3) = w
(-3)+(-3) = -6
What does w equal then? 9.
So! What do we know now? we know that for x^2 + 6x to be a perfect square, we need to add 9 to it. However, we can't just DO that. We wouldn't maintain equality and whatnot. So let's backtrack for a moment. Remember, our original problem was (and I'm not gonna include the y = part to make it easier on myself)
2x^2 - 12x + 23
We then made this into
2 (x^2 - 6x) + 23
We WANT TO have this:
2 (x^2 - 6x +9) + something
so we can factor and have a perfect square. Again, we can't just add 9. HOWEVER. If we were to do this, we would be okay : (and I'll explain why we did this as we go through)
Starting with our original,
2x^2 - 12x + 18 - 18 + 23
All I did was add and subtract 18 from the problem. because 18 -18 is 0, I haven't changed the value of anything. But why did I do this? well, what if I take a 2 out of the first three numbers.
2 ( x^2 - 6x + 9) -18 + 23
I have the perfect square we solved for earlier! now I can factor it.
x^2 - 6x + 9 factors to (x-3)(x-3) or (x-3)^2
So, now I have:
2 (x-3)^2 - 18 + 23
I can further simplify this by doing the -18 + 23
y = 2 (x-3)^2 + 5
So, it's now in the format you needed!
Our first step is complete, and honestly that was the hard part haha. Feel free to ask if the above stuff ^^^ doesn't make sense. Let's now move to step two: defining that k, p, and q stuff.
For reference, recall that:
k = vertical stretch
p = horizontal movement
q = vertical movement
To quickly review what each of these means, a vertical stretch is when you "pull" or stretch a curve vertically, or up or down. A horizontal translation or movement is when your curve moves left or right, and a vertical movement is when your curve moves up or down. How do we find those values when faced with a function though? they're actually pretty easy to spot, if you know what your BASIC function is. Your basic function is when you ignore all values except varaibles. If we look at our function:
y = 2 (x-3)^2 + 5
to find our basic function, we need to ignore everything but varaibles. So, the +5 goes away, -3 goes, and the multiplying by 2 goes. We keep the exponent 2 because you can imagine that as saying "x * x", so really that operation DOES involve a variable. So, our basic function is:
y = x^2
Now we can identify p, q, and k. Let's start with k. A vertical stretch can be identified by a value being MULTIPLIED or DIVIDED by the basic function. Our basic function is x^2. 2 is being multiplied by x^2, so it is a vertical stretch. In this case:
k = 2
Now for p! P is a horizontal movement. a horizontal movement occurs if you add or subtract DIRECTLY from x, not the basic function. Let's look back at our function for this one.
y = 2 (x-3)^2 + 5
We know it isn't 2 (that's multiplication) so we're left with -3 and 5. We want something added or subtracted DIRECTLY from x, not the basic function x^2. In this case, it's -3. So,
p = -3
Finally, we have q, which is a vertical movement. Those occur when you add or subtract something from the BASIC FUNCTION. 5 is being added to the basic function, not directly to x, so it's a vertical movement. So,
q = 5
So yeah, that's all there is to that last bit. Please let me know if ANY OF this doesn't make sense. Especially the beginning part. I know it's rough, trust me haha. Anyway, hope this helps!
It's understood that c and d just stand for constants, or regular numbers without the variable x attached. So! How do we do that? First off, we take out a 2 from JUST 2x^2 and -12x. We can write this as
y = 2 (x^2 -6x) + 23
At this point, we need to recall that we want that (x^2 - 6x) part to be a "perfect square" or that when we factor it, the two factors are the same thing. So, if we wanted x^2 - 6x to be a perfect square, what would we theoretically need to add onto it? To further explain this, I'm gonna rewrite the above quadratic equation as this:
x^2 - 6x + w
Where w is the unknown value that will make this quadratic a perfect square. So, what would w have to be to make this a perfect square? well, if this were just a normal quadratic, we'd need to find two numbers that multiply together to give me the coefficient of x^2 × w, which is, in this case, 1×w, or just w. Those same two numbers need to add together to give me -6.
( ) ( ) = w
( ) + ( ) = -6
We also know these two numbers have to be the SAME, so we factor into a perfect square. The only number, when added to itself, that equals -6, is -3. So, our two mystery numbers have to be -3 and -3.
(-3)(-3) = w
(-3)+(-3) = -6
What does w equal then? 9.
So! What do we know now? we know that for x^2 + 6x to be a perfect square, we need to add 9 to it. However, we can't just DO that. We wouldn't maintain equality and whatnot. So let's backtrack for a moment. Remember, our original problem was (and I'm not gonna include the y = part to make it easier on myself)
2x^2 - 12x + 23
We then made this into
2 (x^2 - 6x) + 23
We WANT TO have this:
2 (x^2 - 6x +9) + something
so we can factor and have a perfect square. Again, we can't just add 9. HOWEVER. If we were to do this, we would be okay : (and I'll explain why we did this as we go through)
Starting with our original,
2x^2 - 12x + 18 - 18 + 23
All I did was add and subtract 18 from the problem. because 18 -18 is 0, I haven't changed the value of anything. But why did I do this? well, what if I take a 2 out of the first three numbers.
2 ( x^2 - 6x + 9) -18 + 23
I have the perfect square we solved for earlier! now I can factor it.
x^2 - 6x + 9 factors to (x-3)(x-3) or (x-3)^2
So, now I have:
2 (x-3)^2 - 18 + 23
I can further simplify this by doing the -18 + 23
y = 2 (x-3)^2 + 5
So, it's now in the format you needed!
Our first step is complete, and honestly that was the hard part haha. Feel free to ask if the above stuff ^^^ doesn't make sense. Let's now move to step two: defining that k, p, and q stuff.
For reference, recall that:
k = vertical stretch
p = horizontal movement
q = vertical movement
To quickly review what each of these means, a vertical stretch is when you "pull" or stretch a curve vertically, or up or down. A horizontal translation or movement is when your curve moves left or right, and a vertical movement is when your curve moves up or down. How do we find those values when faced with a function though? they're actually pretty easy to spot, if you know what your BASIC function is. Your basic function is when you ignore all values except varaibles. If we look at our function:
y = 2 (x-3)^2 + 5
to find our basic function, we need to ignore everything but varaibles. So, the +5 goes away, -3 goes, and the multiplying by 2 goes. We keep the exponent 2 because you can imagine that as saying "x * x", so really that operation DOES involve a variable. So, our basic function is:
y = x^2
Now we can identify p, q, and k. Let's start with k. A vertical stretch can be identified by a value being MULTIPLIED or DIVIDED by the basic function. Our basic function is x^2. 2 is being multiplied by x^2, so it is a vertical stretch. In this case:
k = 2
Now for p! P is a horizontal movement. a horizontal movement occurs if you add or subtract DIRECTLY from x, not the basic function. Let's look back at our function for this one.
y = 2 (x-3)^2 + 5
We know it isn't 2 (that's multiplication) so we're left with -3 and 5. We want something added or subtracted DIRECTLY from x, not the basic function x^2. In this case, it's -3. So,
p = -3
Finally, we have q, which is a vertical movement. Those occur when you add or subtract something from the BASIC FUNCTION. 5 is being added to the basic function, not directly to x, so it's a vertical movement. So,
q = 5
So yeah, that's all there is to that last bit. Please let me know if ANY OF this doesn't make sense. Especially the beginning part. I know it's rough, trust me haha. Anyway, hope this helps!
