H haley2011 New member Joined Apr 4, 2010 Messages 2 Apr 4, 2010 #1 Find the equation of the tangent line to g(x)=square root 2x+9 at the point where x=0. Give answer in slope-intercept form. I have g(x)=(2x+9)^1/2 g'(x)=1/2(2x+9)^-1/2 =1 over (2x+9)^1/2 I don't know where to go from there.
Find the equation of the tangent line to g(x)=square root 2x+9 at the point where x=0. Give answer in slope-intercept form. I have g(x)=(2x+9)^1/2 g'(x)=1/2(2x+9)^-1/2 =1 over (2x+9)^1/2 I don't know where to go from there.
D Deleted member 4993 Guest Apr 4, 2010 #2 haley2011 said: Find the equation of the tangent line to g(x)=square root 2x+9 at the point where x=0. Give answer in slope-intercept form. I have g(x)=(2x+9)^1/2 g'(x)=1/2(2x+9)^-1/2 =1 over (2x+9)^1/2 I don't know where to go from there. Click to expand... What is the slope of the tangent line? What is the point of tangency (i.e. the point where the tangent touches the curve)? What is the equation of a straight-line when you know the slope and a point on the line?
haley2011 said: Find the equation of the tangent line to g(x)=square root 2x+9 at the point where x=0. Give answer in slope-intercept form. I have g(x)=(2x+9)^1/2 g'(x)=1/2(2x+9)^-1/2 =1 over (2x+9)^1/2 I don't know where to go from there. Click to expand... What is the slope of the tangent line? What is the point of tangency (i.e. the point where the tangent touches the curve)? What is the equation of a straight-line when you know the slope and a point on the line?
