Respuesta :
[tex]\bf y=3x^2+5x+4\implies \cfrac{dy}{dx}=6x+5[/tex]
now, we know that derivative will give us the slope, so long we have an x-value.
a line with an y-intercept at 1, will have a point of (0,1), since that's the y-intercept at 1, so let's plug that point and slope in the point-slope form.
[tex]\bf \stackrel{y-intercept}{(\stackrel{x_1}{0}~,~\stackrel{y_1}{1})}~\hspace{10em} slope = m\implies 6x+5 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-1=(6x+5)(x-0) \\\\\\ y-1=6x^2+5x\implies y=6x^2+5x+1[/tex]
since "y" is that equivalent at (0,1) then, let's plug that in our original equation, and solve for "x" to see what our "x" might be.
[tex]\bf y=3x^2+5x+4\implies 6x^2+5x+1=3x^2+5x+4\implies 3x^2=3 \\\\\\ x^2=\cfrac{3}{3}\implies x^2=1\implies x=\pm\sqrt{1}\implies \boxed{x=\pm 1} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{dy}{dx}=6x+5\implies \cfrac{dy}{dx}=6(\pm 1)+5\implies \cfrac{dy}{dx}= \begin{cases} 11\\ -1 \end{cases}[/tex]
