The equation of the line tangent to [tex]f(x) = \sqrt{2x^2 + 3x + 2}[/tex] at x = 0 is given by:
[tex]y = \frac{3\sqrt{2}}{4}x + \sqrt{2}[/tex]
What is a linear function?
A linear function is modeled by:
y = mx + b
In which:
- m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
- b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.
The slope in this problem is the derivative of f(x) at x = 0, as we want the tangent line at x = 0, hence:
[tex]f^{\prime}(x) = \frac{4x + 3}{2\sqrt{2x^2 + 3x + 2}}[/tex]
[tex]f^{\prime}(0) = \frac{3}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{4}[/tex]
Then:
[tex]y = \frac{3\sqrt{2}}{4}x + b[/tex]
The y-intercept is given by f(0), hence:
[tex]f(0) = \sqrt{2 \times 0^2 + 3 \times 0 + 2} = \sqrt{2}[/tex]
Hence the equation is:
[tex]y = \frac{3\sqrt{2}}{4}x + \sqrt{2}[/tex]
More can be learned about linear equations at https://brainly.com/question/24808124
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