A curve has equation y=x^2+2x+5. Write down the coordinates of the minimum point of the curve. Also, show the value of the intercept on the y-axis.
The curve y = x^2 + 2x + 5.
The derivative `dy/dx` = y' = 2x + 2
y'' = 2
At the point of minimum value if x = c,
y' = 0 and y'' at x = c should be positive
y' = 0
=> 2x + 2 = 0
=> x = -1
The value of y at x = -1 is 1 - 2 + 5 = 4
The intercept of y = x^2 + 2x + 5 on the y=axis is y = 4.
The minimum point of y = x^2 + 2x + 5 is (-1, 5) and the y-intercept is 4.