Simplify theequationand keep it inpoint-slope form. y−2=23⋅(x−1)y-2=23⋅(x-1) Solve foryy. Tap for more steps... Simplify23⋅(x-1). y-2=0+0+23⋅(x-1) y-2=23⋅(x-1) y-2=23x+23⋅-1 23andx.
Move the negative in front of the fraction.Step 2 Use the slope and a given point to substitute for and in the point-slope form , which is derived from the slope equation .Step 3 Simplify the equation and keep it in point-slope form.Step...
Find the point of tangency. Since x=2, we evaluate f(2). f(2)=23=8 The point is (2,8). Step 2 Find the value of the derivative at x=2. f′(x)=3x2⟶f′(2)=3(22)=12 The the slope of the tangent line is m=12. Step 3 Find the point-slope form of ...
题目Find the equation of the line in point-slope form, and state the meaning of the slope in context-what information is the slope giving us? 相关知识点: 试题来源: 解析 y- 10=(35)2(x- 12); Every 2 in. of rainfall increases the number of cattle raised per acre by 35.反馈 收藏 ...
To find an equation that is parallel, the slopes must be equal. Find the parallelline using the point-slopeformula. Use the slope( -1/4) and a given point( (4,-2)) to substitute for ( x_1) and ( y_1) in the point-slope form( y-y_1=m(x-x_1)), which is d...
Slope-Intercept Form Lesson Summary Additional Activities Slope Problems Problem 1: Find an equation of the line through the point (5,2) parallel to the line 4x +6y +5 = 0 Problem 2: Show that the lines 2x + 3y = 1 and 6x -4y -1 = 0 are perpendicular. Problem 3: a) Expres...
Point Slope Form: y − 3 = 2(x - 1) Standard Form: 2x - 1y = -1 Angle, Distance, & Intercepts: Angle (θ): 63.435° Distance: 2.236 Δx: 1 Δy: 2 x-Intercept: -0.5 y-Intercept: 1 Steps to Find Slope Start with the slope formula ...
Find the slope of the tangent line at the point {eq}(1,2) {/eq} for the ellipse {eq}2x^2+3y^2=14 {/eq}. Find {eq}y'(1) {/eq}.Tangent Line:We will find the equation of the tangent line by using the point-slope form where we will fi...
Equation of the Tangent Line: We have a function which is the square root of the quadratic function. We will find the derivative of the function to get the slope. Then we will write the equation of the tangent in the point-slope form. ...
Find an equation of the line that passes through the origin and is parallel to the line joining the points (4,5) and (6,12).Point-Slope Form:Given its slope {eq}m {/eq} and a point from the line {eq}(x_0,y_0) {/eq}, the ...