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STEP Support Programme

Assignment 2, q3

I completed part one fine, but had absolutely no clue how to do the next part. Looking at the hints didn't solve anything. I've seen the answers and I have no idea how anyone got to them. Is there a video walkthrough or something like that?

Hi, I found this one tricky too, and I encourage you to keep at it, in the hope that you have as much fun with it as I (eventually) had - I can't provide a video but I can write what lead me to the answer : certainly I couldn't have done it without carefully reviewing my answers to the warm-up and the preparation. Then it was a case of converting the equation to the form $ y = M(x+N)^2 + P $ and sketching all the possibilities of where the bottom of the parabola could be with respect to the range $ -10 \leq x \leq 10 $ : when is it outside or inside, left or right, in the middle. Then I had to do a bit of investigating by sketching what different values of N and P do the the position of the parabola to get a good feel for how it behaves, and then I could go back and solve the problem one situation at a time. I think I filled out about 5 or 6 pages of paper with sketches and equations just to figure things out, but it is certainly a rewarding problem that was well worth the few hours I spent on it. All the best.

I've found one of the answers from just completing the square, but I still have no idea how to show any of this is true. I don't know what to sketch. There are an infinite number of values I could sub in.

It may sound like I'm asking for answers, but I have honestly no idea how to solve this.

I mean, I solved the warm down pretty fine, but this? I don't even know what to put for the working out. I've got to completing the square and that's it. I don't know where to go from there.

Found a solution on thestudentroom. The guy does it algebraically and I sort of understand. Sort of.

hi, can you give me the title of the solution because when I went to thestudentroom, I found nothing. Thank you!

Okay, I didn't break through it myself. I had to look at the answers and a bit of the walkthrough someone else put on the Student Room.

I've understood what's going on. I still consider it an evil question, but I know what's happening now and I know how I'd get the answers. I'd struggle with a similar question, but with time I'd be able to do it.

However, there is one little thing I do not quite understand fully.

The least value for the interval of -10 to 10 is a - ((b^2)/4c). The walkthrough said that this is because in the interval the gradient of the line (2cx + b) changed sign. I don't quite understand why this is so. Could you give an explanation?

Congratulations on your persistence and your research, they are bearing fruit.

For your last question, I think the walk through missed out the "squared" sign : $ (2cx + b)^2 $ because without it that function is just a straight line and so a constant gradient (like your answers to Q2.i, Q2.ii & Q2.iii)

However observe the sketches of the next three parts of Q2, especially Q2.iv and Q2.vi where you can see the line goes down then back up : it's gradient is negative (line going down) on the left, and goes positive (line going up) on the right. In the middle the gradient changes sign, right at the bottom of the bowl in the parabola.

Now look at the x value for the bottom of the bowl in those two questions and compare that to the number in the squared term $ (x + a)^2 $ in each case. It should then be clear that the lowest point on the curve is when $ (x + a)^2 $ is zero. So that will lead you to an answer. It will also lead you to another answer if you consider What happens when à has a negative value, which is what Q2.vii encourages you to do.

Then there are two sketches like Q2.v where the bowl of the parabola is outside the range (if you haven't done it yet, try sketching Q2.v again but for a few bigger values of x than asked for in the question, like say $x=8$ and then consider what would happen if instead of a minus in the question there had been a plus $y=(x+3)^2$

In the end I had 8 sketches in my answer, including the three from Q3.i and my draft had a bunch of exploratory sketches for all sorts of varieties like $y=x^2+1$ $y=x^2-1$ $y=3(x^2)$ $y=-3(x^2)$ etc just so I'd understand what's happening and why.

Hope some of this was helpful.

The 2cx + b wasn't the function, it was the gradient. The guy who did it did it in more of an algebraic way; he differentiated the function and found if it was decreasing or increasing at certain x values and therefore deduced the general notation for each greatest/least value.
Thanks for the extra help, I prefer doing things with algebra than with graphs, so after I've finished the Foundation course, I'll be revisiting this question, just so I know that I can do it.

You said you prefer doing thing with algebra rather than with graphs. I would add that it's better to remain flexible when it comes to solving problems and not only try what you're most comfortable with. Some problems are much more amenable to being solved by diagram/sketch/picture. Indeed, one of the most common phrases my director of studies said to me in first year supervisions (tutorials) was "draw a picture."

I would echo the comment that sketching a few graphs is probably the best way to tackle this question.

The answers are different depending on where the vertex of the graph is, and I would start by drawing 4 general graphs, one where the $x$ coordinate of the vertex is less than -10, one where it is between -10 and 0, one where it is between 0 and 10 and one where it is greater than 10. This will cover all the cases. You will also need to complete the square to find the coordinates of the vertex in terms of $a$, $b$ and $c$.

Hi, I had a go at this question too and I was wondering why Case 1 is when v ≤ -10 but Case 2 is when -10 ≤ v ≤ 0. Doesn't this mean that the two cases overlap (when v = -10) and is this allowed when trying to consider different cases?

In this case the min when $v=-10$ is the same when you used either $100c-10b+a$ or if you use $a-\frac {b^2}{4c}$ as when $v=-10$ we have $\frac b {2c}=10$ and substituting this means that the minimum can be written as $a-5b$ in both cases. Hence $v=-10$ can be included in both parts.

However it would also be equally valid (and possibly preferable) to exclude $v=-10$ from one or other of the cases.