Q) The equation of a curve is y=(x+4)^2+7. Find the co-ordinates of the turning point
Full Explanation:We have our Equation: y=(x+4)2+7
At the turning points, the gradient, (m) is zero m=0 = dy = 0To work out the gradient, we must differentiate the equation dx
The formula to differentiate is: Formula: y=axn Differentiation Formula: dy = naxn-1How to Differentiate: dx 1) We bring the power down and times by the number before the x 2) We take 1 off the power
So lets start with the original Equation y= (x+4)2 + 7
1) We bring the n value down and times with the a value 2x1(x+4)2n= power numbera= number in front of the bracket
2) Now minus 1 from the indice 2(x+4)2-1n= value after the losing bracket 2(x+4)1
And as 7 is a constant, it disappears dy = 2x+8 dx
At the Turning Points, the gradient is zero, therefore dy = 0 dxSo 2x+8= 0Simplify Minus 8 from both side 2x= -8 Divide both sides by 2 x= -4
Substitute x= -4 into Original Equation y= (x+4)2 + 7 y= (-4+4)2 + 7Simplify y= (0)2 + 7 y= 7Concluding Statement: Therefore, the turning point on the curve y= (x+4)2 + 7 are (-4, 7)(Underline final answer)
Concise: y= (x+4)2 + 7 dy = 2(x+4) dx = 2x+8
At Turning Points, dy = 0 dx2x+8= 02x= -8x= -4
Sub x=-4 in y= (x+4)2 + 7y= (-4+4)2+7y= 7
The turning point of the curve y= (x+4)2 + 7 are (-4, 7)
Alternative Way (For easy equations):y= (x+4)2+7For the x-value, take the piece in between the brackets x+4Now take the number and flip the sign -4This is our x-value (-4, ?)
Now take the number outside the brackets (the c) 7This is our y- value y=7
The turning point of the curve is (-4, 7)
(This technique can only be used for simple quadratic equations, usually indicated by one or two marks/ space or no space for working out)
