Final answer:
To find the equation of the tangent line and normal line to the curve at the point (0, 5), take the derivative of the curve, find the slope of the tangent line, and plug in the coordinates of the point to get the equations for the tangent and normal lines.
Step-by-step explanation:
To find the equation of the tangent line and normal line to the curve at the point (0, 5), we need to find the derivative of the curve at that point. The equation of the curve is y = x⁴ - 5eˣ. Taking the derivative, we get dy/dx = 4x³ - 5eˣ. Plugging in the x-coordinate of the point (0), we find that the slope of the tangent line is 4(0)³ - 5e⁰ = -5.
Solving for the y-intercept, we substitute the coordinates of the point (0, 5) into the equation y = mx + b, where m is the slope (-5) and b is the y-intercept. We get 5 = -5(0) + b, which simplifies to b = 5. Therefore, the equation of the tangent line is y = -5x + 5.
The slope of the normal line is the negative reciprocal of the slope of the tangent line. So the slope of the normal line is 1/5. Using the point-slope form of a line, y - y₁ = m(x - x₁), where (x₁, y₁) is the point (0, 5) and m is 1/5, we can derive the equation of the normal line. Simplifying the equation, we obtain y = (1/5)x + 5/5, which simplifies further to y = (1/5)x + 1.