Answer:
The equation of the line is [tex]y = -4\cdot x -2[/tex].
Explanation:
Let the function be [tex]f(x) = 2\cdot x^{2}[/tex] and the line be [tex]4\cdot x + y +3 = 0[/tex]. First, we transform the equation of the line into explicit form:
[tex]y = -4\cdot x + 3[/tex] (1)
By Differential Calculus, the slope at any point of the function is represented by its first derivative, that is:
[tex]f'(x) = 4\cdot x[/tex] (2)
If the line tangent to the function must be parallel to [tex]y = -4\cdot x + 3[/tex], then [tex]f'(x) = -4[/tex]. In consequence, we clear [tex]x[/tex] in (2):
[tex]4\cdot x = -4[/tex]
[tex]x = -1[/tex]
Then, we evaluate the function at the result found above to determine the associated value of [tex]y = f(x)[/tex]:
[tex]f(-1) = 2\cdot (-1)^{2}[/tex]
[tex]y = f(-1) = 2[/tex]
By Analytical Geometry we know that an equation of the line can be formed by knowing both slope ([tex]m[/tex]) and y-intercept ([tex]b[/tex]). If we know that [tex]x = -1[/tex], [tex]y = 2[/tex] and [tex]m = -4[/tex], then the y-intercept of the equation of the line is:
[tex]b = y -m\cdot x[/tex]
[tex]b = 2-(-4)\cdot (-1)[/tex]
[tex]b = 2 -4[/tex]
[tex]b = -2[/tex]
Based on information found previously and the equation of the line, we have the equation of the line that is tanget to the graph of [tex]f(x)[/tex], so that it is parallel to [tex]4\cdot x + y +3 = 0[/tex] is:
[tex]y = -4\cdot x -2[/tex]
The equation of the line is [tex]y = -4\cdot x -2[/tex].
