The tangent to the graph of the function f(x)=x^2+4x+2 is perpendicular to y axis . Determine the point of tangency.

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neela | High School Teacher | (Level 3) Valedictorian

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f(x) = x^2+4x+2.

The slope of the tangent is dy/dx.

Therefore we differentiate the given function y= f(x) = x^2+4x+2.

dy/dx = d/dx {x^2+4x+2}

dy/dx = 2x+4. This slope is of 2x+4 should be perpendicular to y axis or parallel to x axis whose slope is zero.

2x+4 = 0

2x = -4.

x = -2.

At x = -2, the y cordinate could be found by putting x = -2 in the given curve y = x^2+4x+2

y = (2)^2 +4(-2)+2

y = 4-8+2 = -2.

Therefore at (x ,y) = (-2,-2) the tangent to the curve is perpendicular to y axis.

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william1941 | College Teacher | (Level 3) Valedictorian

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Now we know that the tangent to the curve is perpendicular to the y- axis. Therefore it has a slope of 0.

Also the slope can be found by finding the first derivative of f(x) = x^2 + 4x +2

f'(x) = 2x +4

this is equal to 0

=> 2x + 4 =0

=> x = -4/2

=> x = -2

Now substitute x = -2 in f(x) = x^2 + 4x +2

=> (-2)^2 -4*2 +2 = -2

Therefore the required point is (-2 , -2)

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The tangent line to the graph of f(x) is perpendicular to the y axis when it is parallel to x axis.

That means that the tangent line passes through the vertex of the function  f(x) = x^2+4x+2.

We'll calculate the vertex of the parabola, using the coordinates xV and yV.

xV = -b/2a

yV = -delta/4a

We'll identify the coefficients a,b,c.

a = 1

b = 4

c = 2

We'll substitute the coefficients into the coordinates of the vertex:

xV = -4/2*1

xV = -2

yV = (4ac - b^2)/4a

yV = (8 - 16)/4

yV = -8/4

yV = -2

The tangent line, perpendicular to y-axis, is passing through the vertex of the parabola: V(-2 , -2).

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