isn't the minimum point completing the square so (x+ 4)^2 + 15 which becomes (x+4)^2 -1 , therefore the Min value is -1 ??? No point differentiating it. This type of question comes up in a lot of past papers and this is usually what they ask you to do when finding the min/max value.
how do i find the minimum/maximum point of this equation:
x2-8x+15
the 2 being the squared
Firstly, you will need to find gradient of the equation by calculating the derivative. (X^2)-8X+15....... dy/dx = 2x-8 To calculate the minima or maxima, use the calculated gradient. set the equation equal to zero.. 2X-8 = 0. Making X as the subject, 2X-8=0 2X=8 X=8/2 X=4 as the gradient changes from -ve to +ve, for example the point reached is 4 and is positive, so you will have a minimum at x= 4. Using the X value, substituting X=4 into y= (x^2)-8X+15 you will get y= -1. therefore the minimum point is (4, -1) . sorry I have plotted the graph wrong, but I hope it will give you an Idea.