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