My attempt at manipulating the equation $\displaystyle y^2=e^{ax+4}$ $\displaystyle 2lny=(-ax+4)lne$ $\displaystyle lny=\frac{(-ax+4)}{2}$ $\displaystyle lny=\frac{-a}{2}x+2$
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Originally Posted by Punch My attempt at manipulating the equation $\displaystyle y^2=e^{ax+4}$ $\displaystyle 2lny=(-ax+4)lne$ $\displaystyle lny=\frac{(-ax+4)}{2}$ $\displaystyle lny=\frac{-a}{2}x+2$ hi your equation is correct . gradient$\displaystyle =-\frac{a}{2}=\frac{-4-b}{2}$ -- equation 1 the y-intercept is 2 , so we know another point (0,2) gradient$\displaystyle =-\frac{a}{2}=\frac{b-2}{2}$ -- equation 2 so now you hv 2 equations , whats next ?
You are right Using the point (4,-4) gives you the value for a $\displaystyle -4 = -\frac{a}{2} \times 4 + 2$ Then using the point (2,b) gives you the value for b
From $\displaystyle lny=\frac{-a}{2}x+2$, substituting the point (4,-4) I would have to take $\displaystyle lny=ln(-4)$ which shows error 2 in the calculator. confused
Originally Posted by Punch From $\displaystyle lny=\frac{-a}{2}x+2$, substituting the point (4,-4) I would have to take $\displaystyle lny=ln(-4)$ which shows error 2 in the calculator. confused ln y (the whole thing) is -4 not the y inside the ln .
Oh... This means, $\displaystyle lny=\frac{-a}{2}x+2$ is expressed as Y=mX+C where where Y is the y coordinate and X is the x-coordinate
Originally Posted by Punch Oh... This means, $\displaystyle lny=\frac{-a}{2}x+2$ is expressed as Y=mX+C where where Y is the y coordinate and X is the x-coordinate yes
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