Answer:
[tex]y=\dfrac{3x}{2\sqrt 2}+\sqrt 2[/tex]
Step-by-step explanation:
[tex]\text{Given that,}\\\\y = f(x)= \sqrt{2x^2 +3x +2 }\\\\\text{Plug x = 0 in the equation,}\\\\y = \sqrt{2 \cdot 0^2 +3 \cdot 0 +2} =\sqrt{2}\\ \\\text{So, the coordinate is~} \left(0,\sqrt 2 \right).\\\\\text{Now,}\\\\\text{Slope} = f'(x)\\\\~~~~~~~~=\dfrac{1}{2\sqrt{2x^2 +3x +2 }} \cdot \left(4x+3+0\right)\\\\\\~~~~~~~=\dfrac{4x+3}{2\sqrt{2x^2 +3x +2}}\\\\\\[/tex]
[tex]~~~~~=\dfrac{4\cdot 0 +3 }{2 \sqrt{ 2 \cdot 0^2 +3 \cdot 0 + 2}}~~~~~~~~~~~~~~;[\text{at x = 0}]\\\\\\~~~~~~=\dfrac{3}{2\sqrt 2}~~~~~~~~~~~~~~[/tex]
[tex]\text{Equation of tangent line,}\\\\~~~~~~~y-y_1 = f'(x) (x-x_1)\\\\\\\implies y - \sqrt 2 = \dfrac 3{2 \sqrt 2} \left(x - 0\right)\\\\\\\implies y = \dfrac{3x}{2\sqrt 2}+\sqrt 2\\[/tex]
