find the slope of the tangent line to the parabola y=x^2+3x at the point (-1, -2)
using lim x-->a (f(x)-f(a))/x-a
just plug in your function
f(x)-f(a) = (x^2+3x)-(a^2+3a) = (x^2-a^2) + 3(x-a)
= (x+a)(x-a)+3(x-a) = (x-a)/(x+a+3)
so (f(x)-f(a)/(x-a) = x+a+3 for any value x≠a
Now take the limit as x→a
To find the slope of the tangent line to the parabola y = x^2 + 3x at the point (-1, -2) using the limit definition of the derivative, we need to compute the expression lim x->a (f(x) - f(a)) / (x - a), where a is the x-coordinate of the point (-1, -2).
Step 1: Substitute the function f(x) = x^2 + 3x into the expression lim x->a (f(x) - f(a)) / (x - a):
lim x->a [(x^2 + 3x) - (a^2 + 3a)] / (x - a)
Step 2: Simplify the numerator:
lim x->a (x^2 + 3x - a^2 - 3a) / (x - a)
Step 3: Factorize the numerator:
lim x->a [(x - a)(x + a + 3)] / (x - a)
Step 4: Cancel out the common factor (x - a) in the numerator and denominator:
lim x->a (x + a + 3)
Step 5: Now substitute x with -1 to find the value of the limit at the point (-1, -2):
lim x->(-1) (-1 + a + 3)
= (-1 + a + 3)
= (2 + a)
Therefore, the slope of the tangent line to the parabola y = x^2 + 3x at the point (-1, -2) is represented by the expression 2 + a.