# At what point on the curve y = 3x^2 + 3x +7 is the tangent drawn perpendicular to the y axis.

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### 3 Answers

**The tangent drawn 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 y = 3x^2 + 3x +7.**

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 = 3

b = 3

c = 7

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

xV = -3/2*3

**xV = -1/2**

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

yV = (84 - 9)/6

yV = 75/6

**yV = 25/2**

**The perpendicular line to y-axis is passing through the vertex of the parabola: V(-1/2 , 25/2).**

y = 3x^2+3x+7.

The slope of the tangent to the curve at (x,y) is dy/dx.

Therefore we find dy/dx.

dy/dx = (3x^2+3x+7)'

dy/dx = 3*2x+3

dy/dx = 6x+3 . If this should be perpendicular to y axis . So the slope is zero.

Therefore 6x+3 = 0. Or x = -3/6 = -1/2.

So at x= -1/2 , y = (23(-1/2)^2 +3(-1/2)+7 = 3/4-3/2+7 = 6.25.

So the tangent at (-1/2 , 6.25) is Parallel to x axis and perpendicular to y axis.

We are given the curve y = 3x^2 + 3x +7.

Now the slope of a line perpendicular to the y axis is 0.

Also slope is given by the derivative of the curve.

So we differentiate y = 3x^2 + 3x +7

=> y’ = 6x + 3

Now if the slope is 0

=> y’ = 6x +3 =0

=> x = -3/6

=> x = -1/2

Therefore y = 3x^2 + 3x +7 = 3*(-1/2) ^2 + 3*(-1/2) +7 = 6.25

**Therefore the point on the curve where the tangent is perpendicular to the y-axis is (-0.5, 6.25)**