# Find an equation of either the tangent line or normal line?

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1)The tangent line to the curve y=x^3-4 at the point (1,-4)
2)The curve y=ax^2+bx+c passes through the point (1,2) and is tangent to the line y=x at the origin. Find a,b,c.
3)Find an equation of the line tangent to the curve y=3x^2-4x and parallel to the line 2x-y+3=0
4)Find an equation of the line tangent to the curve y=x^4-6x and perpendicular to the line x-2y+6=0

1)The tangent line to the curve y=x3- 4 at the point (1,-4)

The  tangent   formula y - y₁ =m(x - x₁)  where  m =dy/dx

y=x34

dy/dx=3x2 at  the point (1,-4)

m=dy/dx=3(12)

m=3

Substitute m=3  and  the  point (1,-4)in the tangent  formula

y-(-4)=3(x-1)

y+4=3x - 3

Subtract y from each side.

4=3x - y - 3

Subtract 4 from each side.

3x - y - 3 - 4=0

3x - y - 7=0.

The curve y = x3- 4 and the point (1,-4).

There is no tangent line to the curve y = x3- 4 at the point (1,-4), because the point (1, - 4) does not satisfy the curve y = x3- 4.

since   curve  passes  through  point   (1,2)

put x=1    y=2   in  the   curve  equation  y=ax2 +bx+c

then we get

2=a12 +b1+c

a+b+c=2-------------.>(1)

slope  of line  y=x,    m=1------------>(2)

dy/dx=2ax+b

at  origin(0,0)dy/dx=2a(0)+b

since  dy/dx=m=b=2

since  tangent  touches curve  at origin(0,0)  is also  point   on   curve

put  x=0 y=0 in   equation y = ax2+bx+c

then  we  get  c=o

put  b,c  values  in  equation(1)

a+2+0=0

a= -2

therefore  curve  eqation  y=  -2x2-2x

a= -2  b=2  c=0

The values are  a = 1, b = 1, c = 0.

3) Equation of curve y=3x2- 4x

slope of given line = [-x coefficient]/[y coefficient]= -2/-1  =2

dy/dx = 6x - 4

6x - 4 = 2 (The curve parallel to the line so slopes are equal)

6x - 4 + 4 = 2 + 4

6x = 6

Divide each side by 6.

x=1

substitute x = 1 in the curve.

y=3(12)-4(1).

y=3-4

y=-1

point (1,-1), slope =2

The equation of line(y-(-1) )=2(x-1)

(y+1)=2x-2

y+1=2x - 2

1=2x - y - 2

Subtract 1 from each side.

2x - y - 3=0.

Therefore the tangent line equation is 2x - y - 3 = 0.

4) Equation of curve y=x4- 6x

slope of given line = [-x coefficient]/[y coefficient]= -1/-2=1/2

m = dy/dx = 4x3-6

4x3-6= -2 (The curve perpendicular to the line so slopes are m=-1/m)

4x3-6+6  = -2+6

4x3 = 4

Divide each side by 4.

x3 = 1 then x = 1.

substitute x = 1 in the curve.

y=(1)4-6(1)

y=1-6

y=-5

point (1,-5), slope =-2

The equation of line(y-(-5) )=-2(x-1)

(y+5)=-2x+2

y+5=-2x+2

2x+y+5=2

Subtract 5 from each side.

2x +y=2 - 5

2x+y =-3

2x+y+3=0

Therefore the tangent perpendicular line equation is 2x +y+3 = 0.

2)The line y = x

Compare it to slope intercept form of line y = mx + c

Slope of the line (m) = 1

• The curve y = ax2 + bx + c ___________ (1)

The curve passes through the point (1, 2)

Substitute x = 1 and y = 2 in  equation (1).

2 = a(1)2 + b(1) + c

2 = a + b + c ______ (2)

• The derivative of the curve is equal to the slope of the tangent line.

y = ax2 + bx + c

dy/dx = 2ax + b

m = 2ax + b

y = ax2 + bx + c tangent to the line y = x at the origin.

m = 2a (0)+ b

m = b

Slope of the line m = b = 1

Since the tangent line touches curve at origin (0, 0) also point on a curve.

Substitute x = 0 and y = 0 in curve.

0 = a(0)2 + b(0) + c

c = 0

b = 1 and c = 0 in a + b + c = 2

a + 1 + 0 = 2

a = 2 - 1

a = 1

Substitute the a ,b , c alues in the equation (2).

The curve is y = x2 + x

The values are a = 1, b = 1, c = 0.