# SOLVED: Solve the following problems. Show your complete solutions. (This is all about Basic Calculus: Slope of the Tangent Line to a Curve). a. Find the equations of the tangent and normal lines to t

We are going to solve the following problems: a you’ll find the equations of the tangent and normal lines to the curve y equals x, squared plus 2 x, plus 3 at x equals negative 1 part. We find the equation of tangent line to the curve y equals x, squared minus 2 x minus 3. That is parallel to the line x, plus y equals 1 pace. We find an equation of tangent line to the curve y equals x, squared minus 3 x plus 4, that is perpendicular to the line x. Minus y equals negative 1 point. So let’s do part a first, so we want the 2 lines. The tangled line and normal line- so we know the slope of the tangent line to the curve at a given value of x- is the derivative of that function, defining the curve evaluated at the value of x. So let’s say we have the derivative of y with respect to x is this case is a polynomial function of degree. 2 is derivative is 2 x, minus plus 22 x, plus 2, and so the derivative of y, with respect to x, evaluated at x, equals negative 1 is 2 times negative, 1 plus 2, and that is negative, 2 plus 2, and that is 0. And this is the slope of the tangent line to the curve at x equals negative 1 point. So it means that this standard line is horizontal because it’s slope is 0, so the tangent line to the curve at negative 1 is horizontal, so its equation is of type y, equal, constant. So to determine that constant, we calculate the point of tangent to the point of tangency is obtained by evaluating the curve at negative 1 point. That is, you apply. The formula is y is x, squared plus 2 x plus 3 point, so we get negative 1 square plus 2 times negative 1 plus 3, that is 1 minus 2 plus 3 is 2. So the point of tendency is negative 12, because the first component is negative. 1 and second 1 is the value of y is 2, and we know this function x, squared plus 2 x, plus 3 is a parable. So it means if the tangent line is horizontal, that point of tendency must be the vertex problem and that can be verified easily. This is the case, and so we can say because the tonal line is horizontal than the equation of the tandant line at x. Equals negative 1 is y equals 2 because it’s the value through which passes the tenor line. The point of tangents, okay and now the normal line we know, is perpendicular to the tiny line. So the normal line in this case is a vertical line vertical and sorry the normal line, let’s say better this way since, since the tangent line at negative 1 is horizontal than the normal. At that same point, of course, the normal line to the curve at 81 x, equals 81 is vertical, and since that line, that vertical line passes through the point of a negative 12, the equation is x, equals negative 1 and because this normal line passes through the Point of tangency at 12 point then, then, its equation is x, equals negative 1 equation of the normal line. So in summer this part a we calculate the derivative of the function at negative 1. That gives the slope of the tenant line, because that slope is 0 point. We know the tangent line is a recent line, and so its equation is of the form y equal, constant value, and that constant value is determined by calculating the point of tendency, because the the teninline passes through that point, and we calculate that value evaluating the parabola. At negative 1, the value of x and we got 2, so the point of tangency is negative 12 and then the equation of the tangent line is y. Equal to the normal line is perpendicular to the tandanline, and so since the tenant line is horizontal, the normal line is vertical and its equation is x, equals constant, and since that line passes through the point of tangency negative 12 o the equation is x, equals negative 1 point: so we have the 2 parts in part a so in part b. Now we want to find the equation of the tana line to the curve that is parallel to the line x, plus y equals 1. If 2 lines are parallel, they have the same slope. We know the slope will be given by the slope of this line, and we know the slope of this line is found by simply solving the equation for y and then the coefficient of x will be the slope so first x, plus y equals 1, which is The equation of the line to which the ear the tangled line is parallel to this means that y is 1 minus x. So, given the equation in this form that is solve for y, this slope of this line is negative 1. It is the coefficient of x when we have solved the equation for y good, so we know the slope of the changing line to the curve, okay and so to know, at which point we have a tangent line. That is that has this slope, we calculate the derivative of the functions defined in the curve, that is the derivative of y with respect to x, and that gives us 2 x minus 2, and we know that at any x this is the slope of the tenant Line added to the curve at the given value of x, so we want the slope to be negative 1, so this equation is solved to find the x at which the tangent line is parallel. Okay to the line x, plus y equals…

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