help me with these equations

Question

Consider the graph of the function $f(x)=-x^2+9$

so far I've come up with (a) $y=5x-1$ (b) (1/2,5) i'm just lost on $c$ and what formulas to use can someone confirm these answers correct and help me out?

Answer 1

Your solution for (a) is certainly wrong. We know the slope of the line will be (1-4)/(2-(-1))=-3/3=-1 and the intercept is -3 so the line will be y=-x-3

Answer 2

(a) To find the slope of the secant line, find the vertical distance between the two points by subtracting $y_2=1$ and $y_1=4$ to get -3. Divide that by the horizontal distance $x_2=2 - x_1=-1$ to get -3/3 or -1. To get the y-intercept, we can start at (-1, 4) and move forward along the x axis to (0, 4-1) = (0, 3). Thus, the equation of the secant line is $y=-x+3$.

(b) Now that we know the slope of the secant line, we simply need to find when the slope of the curve is equal to -1. First we use the power rule to find the derivative equation $f'(x)=-2x$, then we set it equal to -1 to get $-1 = -2x \implies x = 1/2.$

(c) Next, we evaluate the function at $\frac{1}{2}$ to get the $y$-value when the slope is -1 to get $\frac{-1}{2}^2+5 \implies \frac{-1}{4}+5 \implies \frac{19}{4}.$ We find our y-intercept by moving back $\frac{1}{2}x$'s on the graph from the point (1/2, 19/4) to (0, (19/4-(-1/2))) = (0, 21/4) making the equation of the tangent line $y=-x + \frac{-21}{4}$.

(d) Well, you probably know enough by now to graph them all yourself.