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Core 1 question: Draw the graph "y = 12 - x - x^2"

I always found drawing graphs one of the hardest parts of maths - especially in C1 where you cannot use a calculator to help you. So, I used some simple rules to find key parts of the graph which would simplify the drawing.

Find the y-intercept (when x = 0)
Find any x-intercepts (when y = 0)
Find what "shape" it should be by finding using the highest powered x.
Find the turning point i.e. the maximum or minimum. By using differentiation.

In this case: Draw the graph "y = 12 - x - x²"

Find y-intercept. Insert "x=0" in the equation. Therefore, y = 12 - 0 - 0 = 12. Therefore y-intercept = 12. (0, 12)
Find x-intercepts. Do this by factorising. " y = (3 - x) (4 + x) " From this you can deduce that the x-intercepts (i.e. when y = 0) are 3 and -4. (3, 0) and (-4, 0)
Find the shape of the curve by using the highest indicie power... the "-x^{2 "} and from this we can decude it is a Negative Quadratic which means the shape it makes is an 'frown'.
Find the turning point. We know from "-x^{2 "} that it is a Maximum we are looking for.

So find the Maximum, we differentiate the original equation "y = 12 - x - x²" (remember differentiating means "times by the original power and reduce the power by 1, e.g. x² becomes 2x). Differentiation is used to find the gradient of a curve at any point.

Therefore, y = 12 - x - x² ... dy/dx = -1 - 2x

Now this is the gradient equation at any point. When there is a Maximum turning point the gradient always = 0. Therefore, to find the co-ordinates for the turning point of this graph, we just set dx/dy = 0.

dx/dy = 0 = -1 - 2x. SOLVE. 1 = - 2x, x = -1/2. Then put x = -1/2 into the y equations and y = 12 1/4. (-1/2, 12 1/4)

So we have the co-ordinates needed to construct the graph. We need to plot the co-ordinates. Then use our knowledge that it is a negative quadratic (so is a frown) and then join up the co-ordinates with a smooth curve.