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secondary 3 | E Maths

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

With Sec 4 A Maths techniques this question becomes straightforward.

I will write up the solutions later.

It should be y = k (3^-x).

y = 3 (3^-0.05x) models this equation better.

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

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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).

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

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