what are the functions of programs h(x) = x^2-2 , l(x)= (x+3)2, j(x) (x+1)^2 +3

Question

what are the slope of and functions of these problems h(x) =x^2 -2  ,  l(x) = (x+3)^2  , j(x) =(x+1)^2 +3

Answer 1

All three equations are quadratic. If you decide to take calculus, you will learn about derivatives which are formulas for slopes of functions. To get the slope at a point P = (x , f(x)), consider another point on the graph, Q = (x+h , f(x+h)) close to P (h is small).

Then the slope at point  P is approximately equal to the slope of line PQ, provided that the function is well behaved. The closer  the point Q is to point P, the more accurate the approximation. Let's use it on a quadratic polynomial f(x) = ax²+bx+c

Then the slope of line PQ is:

m = [f(x+h) - f(x)] / [(x+h) - x]

    = [a(x+h)²+b(x+h)+c - (ax²+bx+c)] / h

    = (ax²+2axh+ah²+bx+bh+c-ax²-bx-c) / h

    = (2axh+ah²+bh) / h

    = 2ax+ah+b

Now if h approaches 0, ah approaches 0 so the slope (derivative) of f(x) is given by 2ax+b.

The limit of the slope formula for line PQ as Q approaches P is called the definition of a derivative. The derivative of f(x) is written as f'(x) using an apostrophe after the function's name.

Let's apply the 2ax+b result on your functions.

h(x) = x²-2 so h'(x) = 2x.

l(x) = (x+3)² = x²+6x+9 so l'(x) = 2x+6

j(x) = (x+1)²+3 = x²+2x+4 so j'(x) = 2x+2

 

Answer 2

Since these functions are all second-order polynomials ("quadratics"), their slope is constantly changing at every point. 

The only way that a function can have the same slope everywhere is if the function is "linear" and theonly exponent of X is one.  (I like to illustrate this by exerting force on each end of a plastic pen, bending it while grunting that "it takes power to bend a line!"   Then, when I release the pen, it becomes a straight line -- just like the exponent of X.  Any power of X, other than one or zero, will create a CURVE.)

The graph for each of these functions is a parabola, and there is only one point where the slope is zero. You need a bit of calculus to find this point, by finding the derivative (an expression for "instantaneous slope"), then setting it equal to zero and solving for X.  Below are the derivatives (written with an apostrophe for "prime) for each of these functions, along with the coordinate of the "extremum" point (where the slope is zero).

    h'(x) =2x,   l'(x) = 2x+6 ,   j'(x) = 2x+2
      (0, -2)        (-3,0)              (-1,3)