Upload a photo of a Singapore homework and someone will email you the solution for free. Question

secondary 3 | E Maths

Anyone can contribute an answer, even non-tutors.

Lee Meimei

chevron_right chevron_right

Sec 3 E maths qn. How to do? Please help thanks.

Date Posted: 1 month ago
Views: 21
Eric Nicholas K
1 month ago
The nice thing about this is that if (0, 3) is a point on the curve, so does (2, 3), since the line of symmetry is x = 1.

With Sec 4 A Maths techniques this question becomes straightforward.

I will write up the solutions later.
Eric Nicholas K
1 month ago
Something is not right with the equation y = k (3^x). The graph equation is wrong.

It should be y = k (3^-x).
J
1 month ago
Even k(3^(-x)) doesn't look close
Eric Nicholas K
1 month ago
That’s the best by far. It’s a “sketch” after all.

y = 3 (3^-0.05x) models this equation better.
J
1 month ago
To fit this equation, yes 3(3^(-x))

To fit the shape, 3^(-0.45x) + 2 looks better. Teachers really got to be careful in setting the graphs.

Also, If the assumption that (1,4) isn't minimum point this question probably is undoable

done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
Here are my workings for this question.

The downward-going graph should not have the equation y = k (3^x) because the graph of this equation is upsloping and fully above the y-axis for positive values of k and downsloping and fully below the y-axis for negative values of k.

A correct equation for such curves is y = k (3^-x).
Date Posted: 1 month ago
Another A Math technique (optional, might be good to know)
c = 3 , k = 3, a + b = 1 (from Eric's above workings)
If the point (1,4) is taken as the maximum point, then it is tangent to the line y = 4. So it only meets that line at that one point.
We have y = 4 and y = ax² + bx + 3
4 = ax² + bx + 3
ax² + bx - 1 = 0
Since line meets the curve at only 1 point, there are real and equal roots. So discriminant (b² - 4ac) = 0
b² - 4a(-1) = 0
b² + 4a = 0 ①
Since a + b = 1, b = 1 - a ②
Sub ② into ①, (1 - a)² + 4a = 0
1 - 2a + a² + 4a = 0
a² + 2a + 1 = 0
(a + 1)² = 0
a + 1 = 0
a = -1
Then b = 1 - (-1) = 1 + 1 = 2
done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
Date Posted: 1 month ago
done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
Here is an alternative approach for the quadratic graph.
Date Posted: 1 month ago
done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
A third approach, using Sec 4 A Maths techniques, purely as an additional approach if you are interested. Not to be used in E Maths papers since only E Maths techniques are permitted in E Maths papers.
Date Posted: 1 month ago 