# 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.

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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.**