Asked by sera
                find the gradient of the tangent to the curve y=3x^2+2x-1 p(1,4)
            
            
        Answers
                    Answered by
            Bosnian
            
    Slope = gradient if tangent line = dy/dx = m
In this case:
dy / dx = m = 3 • 2 x + 2 = 6 x + 2
In point where x = 1
m = 6 x + 2 = 6 • 1 + 2 = 6 + 2 = 8
Point slope equation of straight line is:
y − y1 = m ( x − x1 )
In this case:
x1 = 1 , y1 = 4 , m = 8
so
y − y1 = m ( x − x1 )
y − 4 = 8 ( x − 1 )
y − 4 = 8 x − 8
Add 4 to both sides.
y = 8 x - 4
    
In this case:
dy / dx = m = 3 • 2 x + 2 = 6 x + 2
In point where x = 1
m = 6 x + 2 = 6 • 1 + 2 = 6 + 2 = 8
Point slope equation of straight line is:
y − y1 = m ( x − x1 )
In this case:
x1 = 1 , y1 = 4 , m = 8
so
y − y1 = m ( x − x1 )
y − 4 = 8 ( x − 1 )
y − 4 = 8 x − 8
Add 4 to both sides.
y = 8 x - 4
                    Answered by
            Bosnian
            
    My typo.
Slope = gradient of tangent line = dy/dx = m
    
Slope = gradient of tangent line = dy/dx = m
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