I'm working through the Pearson Edexcel Pure Mathematics Year 1/ AS practice book. I've been over this question twice using different methods, and I really think they've made a mistake. I'd be grateful for anyone else who can cast their eye over my working and see if it's me or them who's mistaken...

The question is "Find the coordinates of the turning point of y = -x^2 - 13x - 42".

Here's one of the methods I used to find the turning point:

y = -x^2 - 13x - 42 --> Multiply everything by -1

y = x^2 + 13x + 42 --> Complete the square

y = (x + 13/2)^2 - 169/4 + 42 --> Put everything outside the bracket /4

y = (x + 13/2)^2 - 169/4 + 168/4 --> Simplify

y = (x + 13/2)^2 - 1/4

--> Turning point is (-13/2, -1/4)

The other method I tried was just to factorise out -1 at the start and put everything else in square brackets, then complete the square as normal.

Both methods returned me coordinates of (-13/2, -1/4).

However the book says the answer should be (-13/2, 1/4).

It doesn't show how they arrived at that answer.

Am I mistaken, or are they?!

The question is "Find the coordinates of the turning point of y = -x^2 - 13x - 42".

Here's one of the methods I used to find the turning point:

y = -x^2 - 13x - 42 --> Multiply everything by -1

y = x^2 + 13x + 42 --> Complete the square

y = (x + 13/2)^2 - 169/4 + 42 --> Put everything outside the bracket /4

y = (x + 13/2)^2 - 169/4 + 168/4 --> Simplify

y = (x + 13/2)^2 - 1/4

--> Turning point is (-13/2, -1/4)

The other method I tried was just to factorise out -1 at the start and put everything else in square brackets, then complete the square as normal.

Both methods returned me coordinates of (-13/2, -1/4).

However the book says the answer should be (-13/2, 1/4).

It doesn't show how they arrived at that answer.

Am I mistaken, or are they?!

Remember when you are completing square, you are working with an expression, not an equation.

The minus one doesn't disappear in the first line.

(i.e. The mistake is the first line doesn't imply the second.)

The minus one doesn't disappear in the first line.

(i.e. The mistake is the first line doesn't imply the second.)

(edited 7 months ago)

Original post by korma

I'm working through the Pearson Edexcel Pure Mathematics Year 1/ AS practice book. I've been over this question twice using different methods, and I really think they've made a mistake. I'd be grateful for anyone else who can cast their eye over my working and see if it's me or them who's mistaken...

The question is "Find the coordinates of the turning point of y = -x^2 - 13x - 42".

Here's one of the methods I used to find the turning point:

y = -x^2 - 13x - 42 --> Multiply everything by -1

y = x^2 + 13x + 42 --> Complete the square

y = (x + 13/2)^2 - 169/4 + 42 --> Put everything outside the bracket /4

y = (x + 13/2)^2 - 169/4 + 168/4 --> Simplify

y = (x + 13/2)^2 - 1/4

--> Turning point is (-13/2, -1/4)

The other method I tried was just to factorise out -1 at the start and put everything else in square brackets, then complete the square as normal.

Both methods returned me coordinates of (-13/2, -1/4).

However the book says the answer should be (-13/2, 1/4).

It doesn't show how they arrived at that answer.

Am I mistaken, or are they?!

The question is "Find the coordinates of the turning point of y = -x^2 - 13x - 42".

Here's one of the methods I used to find the turning point:

y = -x^2 - 13x - 42 --> Multiply everything by -1

y = x^2 + 13x + 42 --> Complete the square

y = (x + 13/2)^2 - 169/4 + 42 --> Put everything outside the bracket /4

y = (x + 13/2)^2 - 169/4 + 168/4 --> Simplify

y = (x + 13/2)^2 - 1/4

--> Turning point is (-13/2, -1/4)

The other method I tried was just to factorise out -1 at the start and put everything else in square brackets, then complete the square as normal.

Both methods returned me coordinates of (-13/2, -1/4).

However the book says the answer should be (-13/2, 1/4).

It doesn't show how they arrived at that answer.

Am I mistaken, or are they?!

You haven't changed the sign of y in the first line. The curve is an inverted parabola ....

Why not differentiate to check?

y = -x^2 - 13x - 42

dy/dx = -2x - 13 = 0 for st point .... x = -13/2 the subst into curve

Original post by tonyiptony

Remember when you are completing square, you are working with an expression, not an equation.

The minus one doesn't disappear in the first line.

(i.e. The mistake is the first line doesn't imply the second.)

The minus one doesn't disappear in the first line.

(i.e. The mistake is the first line doesn't imply the second.)

Ahhhhh okay I think I get you. Thanks so much for explaining. Does this look correct, then?

y = -x^2 - 13x - 42 --> Multiply everything by -1

-y = x^2 + 13x + 42 --> Complete the square

-y = (x + 13/2)^2 - 169/4 + 42 --> Put everything outside the bracket /4

-y = (x + 13/2)^2 - 169/4 + 168/4 --> Simplify

-y = (x + 13/2)^2 - 1/4 --> Multiply everything by -1 again

y = -(x + 13/2)^2 + 1/4

--> Turning point is (-13/2, 1/4)

Thanks again for explaining, hopefully making this mistake now will help me remember when it comes to the exam!

Original post by Muttley79

You haven't changed the sign of y in the first line. The curve is an inverted parabola ....

Why not differentiate to check?

y = -x^2 - 13x - 42

dy/dx = -2x - 13 = 0 for st point .... x = -13/2 the subst into curve

Why not differentiate to check?

y = -x^2 - 13x - 42

dy/dx = -2x - 13 = 0 for st point .... x = -13/2 the subst into curve

Thank you so much for responding and explaining. What you said about me inverting the parabola makes complete sense to me now!

I haven't reached the differentiation part of the course yet but will come back to your comment once I've learned how to do that

Original post by korma

Ahhhhh okay I think I get you. Thanks so much for explaining. Does this look correct, then?

y = -x^2 - 13x - 42 --> Multiply everything by -1

-y = x^2 + 13x + 42 --> Complete the square

-y = (x + 13/2)^2 - 169/4 + 42 --> Put everything outside the bracket /4

-y = (x + 13/2)^2 - 169/4 + 168/4 --> Simplify

-y = (x + 13/2)^2 - 1/4 --> Multiply everything by -1 again

y = -(x + 13/2)^2 + 1/4

--> Turning point is (-13/2, 1/4)

Thanks again for explaining, hopefully making this mistake now will help me remember when it comes to the exam!

y = -x^2 - 13x - 42 --> Multiply everything by -1

-y = x^2 + 13x + 42 --> Complete the square

-y = (x + 13/2)^2 - 169/4 + 42 --> Put everything outside the bracket /4

-y = (x + 13/2)^2 - 169/4 + 168/4 --> Simplify

-y = (x + 13/2)^2 - 1/4 --> Multiply everything by -1 again

y = -(x + 13/2)^2 + 1/4

--> Turning point is (-13/2, 1/4)

Thanks again for explaining, hopefully making this mistake now will help me remember when it comes to the exam!

Looks good. Nice work! I would prefer a "one-line" calculation, so the second step would just be "= -(x^2+13x+42)", i.e. no need to "y=" every time. Or even, I tend to only factor the first 2 terms, but slight variation in prepping for the step of completing square wouldn't matter as long as the arithmetic is correct. Do stick with the procedure you feel the most comfortable with.

Now a minor tidbit if you're interested. This particular problem has a hack.

Notice that the roots of the quadratic is -6 and -7 (as -x^2 - 13x - 42 can be factored as -(x+6)(x+7)).

So by the symmetry of the parabola, the x-coordinate of the turning point is precisely the middle of the two roots, i.e. -13/2.

Of course, if (i) the question specifically requires you to use completing square; or (ii) the quadratic doesn't factor nicely (at least if you can't do it in your head), then this hack doesn't work.

Another tidbit, actually you can write down the x-coordinate of the turning point (i.e. no calculation required), if you happen to know some properties relating the coefficients of terms in a quadratic and its roots.

Also another comment, I often find using calculus to find turning points of quadratics unnecessary. Granted I can complete squares really quickly having done loads... Then again, do what you're comfortable with.

(edited 7 months ago)

Original post by tonyiptony

Now a minor tidbit if you're interested. This particular problem has a hack.

Notice that the roots of the quadratic is -6 and -7 (as -x^2 - 13x - 42 can be factored as -(x+6)(x+7)).

So by the symmetry of the parabola, the x-coordinate of the turning point is precisely the middle of the two roots, i.e. -13/2.

Of course, if (i) the question specifically requires you to use completing square; or (ii) the quadratic doesn't factor nicely (at least if you can't do it in your head), then this hack doesn't work.

Another tidbit, actually you can write down the x-coordinate of the turning point (i.e. no calculation required), if you happen to know some properties relating the coefficients of terms in a quadratic and its roots.

Notice that the roots of the quadratic is -6 and -7 (as -x^2 - 13x - 42 can be factored as -(x+6)(x+7)).

So by the symmetry of the parabola, the x-coordinate of the turning point is precisely the middle of the two roots, i.e. -13/2.

Of course, if (i) the question specifically requires you to use completing square; or (ii) the quadratic doesn't factor nicely (at least if you can't do it in your head), then this hack doesn't work.

Another tidbit, actually you can write down the x-coordinate of the turning point (i.e. no calculation required), if you happen to know some properties relating the coefficients of terms in a quadratic and its roots.

Sorry, but I'm kind of unconvinced by this as a hack. Finding the x-coordinate when completing the square also has no calculation required (or at least, it's *exactly* the same calculation as in your hack), and the algebraic argument for completing the square doesn't involve you needing to find the roots or make a symmetry argument (so to my mind, it's "less work", even though obviously the literal calculation is the same).

Original post by DFranklin

Sorry, but I'm kind of unconvinced by this as a hack. Finding the x-coordinate when completing the square also has no calculation required (or at least, it's *exactly* the same calculation as in your hack), and the algebraic argument for completing the square doesn't involve you needing to find the roots or make a symmetry argument (so to my mind, it's "less work", even though obviously the literal calculation is the same).

Touche.

I guess I'd just mention it because apparently so many people struggle with the algebra of completing square...

Simply another perspective, really. Is it harder? Eh, not for me to judge. Just something the sprung to mind that could very well be useless.

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