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secondary 3 | E Maths
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In this instance, you can perform simultaneous equations on the curve and the line to get an equation in x. We work out way towards the required format.
The value of x, in this case, might not be necessary.
Sub x = -2.45,
(-2.45)³ + A(-2.45) + B = 0
-14.706125 - 2.45A + B = 0
B = 2.45A + 14.706125
This is analogous to y = 2.45x + 14.706125
Unless you have another value of x, it's not quite possible to proceed further.
There is an infinite number of (A,B) pairs that satisfy this equation.
However, if the actual question is like what Eric said above, then you'll have to equate the curve (assuming its quadratic) to the equation of another function that intersects it.
For example,
Let's say you were asked to plot the quadratic curve with equation y = 3x² - 8 and a reciprocal curve with equation y = -25/x on the same piece of graph paper.
Plotting these two graphs will yield one intersection point x ≈ -2.46 (3s.f)
(The best estimation from the graph yields x = -2.45 as it probably lies somewhere in the middle of the small box in the grid of the graph paper i.e between x = -2.4 and x = -2.5, subjected on the scale you use)
Equating the two,
3x² - 8 = -25/x
Multiply both sides by x,
3x³ - 8x = -25
3x³ - 8x + 25 = 0
Divide throughout by 3,
x³ - 8/3 x + 25/3 = 0
A = -8/3 or 2⅔
B = 25/3 or 8⅓
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