</section> <section style="font-size: 50px";> <h2 id="derivatives---equations-of-tangent-and-normals">Derivatives - Equations of Tangent and Normals</h2>

</section> <section style="font-size: 50px";> <h2 id="introduction">Introduction</h2> <ul> <li>In this lesson, we will learn about equations of tangent and normals of a function.</li> <li>The derivative of a function can be used to find the equation of the tangent and normal lines at a point on the curve.</li> <li>Tangent is a line that touches the curve at a specific point.</li> <li>Normal is a line perpendicular to the tangent and passes through the same point.

</section> <section style="font-size: 50px";> <h2 id="equation-of-tangent">Equation of Tangent</h2> <ul> <li>The equation of the tangent line to the curve at a point (x₁, y₁) can be found using the derivative of the function.</li> <li>Let f(x) be the function, and f’(x) be its derivative.</li> <li>The equation of the tangent line is given by: <code>y - y₁ = f'(x₁)(x - x₁)</code></li> <li>Where: <ul> <li>(x₁, y₁) is the point of tangency</li> <li>f’(x₁) is the derivative of the function evaluated at x = x₁

</section> <section style="font-size: 50px";> <h2 id="example-1">Example 1</h2> <ul> <li>Consider the function f(x) = 3x² + 2x - 7.</li> <li>Find the equation of the tangent line at the point (2, -1).</li> <li>Solution: <ul> <li>First, find the derivative of the function: f’(x) = 6x + 2.</li> <li>Substitute x = 2 into the derivative to find f’(2) = 14.</li> <li>The equation of the tangent line is: y - (-1) = 14(x - 2).</li> <li>Simplify the equation: y + 1 = 14x - 28.

</section> <section style="font-size: 50px";> <h2 id="equation-of-normal">Equation of Normal</h2> <ul> <li> <p>The equation of the normal line to the curve at a point (x₁, y₁) can also be found using the derivative of the function.</p> </li> <li> <p>The equation of the normal line is given by: <code>y - y₁ = -1/f'(x₁)(x - x₁)</code></p> </li> <li> <p>Where:</p> <ul> <li>(x₁, y₁) is the point of tangency</li> <li>f’(x₁) is the derivative of the function evaluated at x = x₁</li> </ul> </li> <li> <p>Notice that the slope of the normal line is the negative reciprocal of the slope of the tangent line.

</section> <section style="font-size: 50px";> <h2 id="example-2">Example 2</h2> <ul> <li>Consider the function g(x) = x³ - 2x.</li> <li>Find the equation of the normal line at the point (1, -1/3).</li> <li>Solution: <ul> <li>First, find the derivative of the function: g’(x) = 3x² - 2.</li> <li>Substitute x = 1 into the derivative to find g’(1) = 1.</li> <li>The equation of the normal line is: y - (-1/3) = -1/1(x - 1).</li> <li>Simplify the equation: y + 1/3 = -(x - 1).

</section> <section style="font-size: 50px";> <h2 id="finding-intersection-points">Finding Intersection Points</h2> <ul> <li>We can find the intersection points of two curves or lines by solving their equations.</li> <li>To find the intersection point between two functions, set their equations equal to each other and solve for the x-coordinate.</li> <li>Once the x-coordinate is found, substitute it back into one of the equations to find the y-coordinate.</li> <li>These intersection points can be useful in understanding the behavior of curves and lines.

</section> <section style="font-size: 50px";> <h2 id="example-3">Example 3</h2> <ul> <li>Find the intersection points of the functions f(x) = x² and g(x) = 2x - 3.</li> <li>Solution: <ul> <li>Set the equations equal to each other: x² = 2x - 3.</li> <li>Rearrange the equation: x² - 2x + 3 = 0.</li> <li>This is a quadratic equation, solve using factoring or quadratic formula.</li> <li>The solutions are x = -1 and x = 3.</li> <li>Substitute x = -1 into f(x) to find y = 1.</li> <li>Substitute x = 3 into f(x) to find y = 9.

</section> <section style="font-size: 50px";> <h2 id="summary">Summary</h2> <ul> <li>Equations of tangent and normals can be found using the derivative of a function.</li> <li>The equation of the tangent line is: <code>y - y₁ = f'(x₁)(x - x₁)</code>.</li> <li>The equation of the normal line is: <code>y - y₁ = -1/f'(x₁)(x - x₁)</code>.</li> <li>Intersection points can be found by solving the equations of the functions or lines.</li> <li>Intersection points are useful in understanding the behavior of curves and lines.

</section> <section style="font-size: 50px";> <h2 id="derivatives---equations-of-tangent-and-normals-1">Derivatives - Equations of Tangent and Normals</h2>

</section> <section style="font-size: 50px";> <h2 id="example-4">Example 4</h2> <ul> <li>Consider the function h(x) = x^4 + 2x^3 - 3x^2.</li> <li>Find the equation of the tangent line at the point (1, 0).</li> <li>Solution: <ul> <li>First, find the derivative of the function: h’(x) = 4x^3 + 6x^2 - 6x.</li> <li>Substitute x = 1 into the derivative to find h’(1) = 4 + 6 - 6 = 4.</li> <li>The equation of the tangent line is: y - 0 = 4(x - 1).</li> <li>Simplify the equation: y = 4x - 4.

</section> <section style="font-size: 50px";> <h2 id="example-5">Example 5</h2> <ul> <li>Consider the function k(x) = √x.</li> <li>Find the equation of the tangent line at the point (4, 2).</li> <li>Solution: <ul> <li>First, find the derivative of the function: k’(x) = 1/(2√x).</li> <li>Substitute x = 4 into the derivative to find k’(4) = 1/(2√4) = 1/4.</li> <li>The equation of the tangent line is: y - 2 = (1/4)(x - 4).</li> <li>Simplify the equation: y = (1/4)x + 1/2.

</section> <section style="font-size: 50px";> <h2 id="equation-of-normal-to-a-circle">Equation of Normal to a Circle</h2> <ul> <li>The equation of the normal to a circle at a given point (x₁, y₁) can be found using the equation of the tangent line.</li> <li>For a circle centered at (a, b) with radius r, the equation of the normal line is: <code>(x - x₁) / (a - x₁) = -(y - y₁) / (b - y₁) = -(r² - (x₁ - a)² - (y₁ - b)²) / r</code></li> <li>Where: <ul> <li>(x₁, y₁) is the given point on the circle</li> <li>(a, b) is the center of the circle</li> <li>r is the radius of the circle

</section> <section style="font-size: 50px";> <h2 id="example-6">Example 6</h2> <ul> <li>Consider a circle centered at (3, -2) with radius 5.</li> <li>Find the equation of the normal line at the point (6, 2) on the circle.</li> <li>Solution: <ul> <li>Substitute the given values into the equation of the normal line.</li> <li>Substitute x₁ = 6, y₁ = 2, a = 3, b = -2, and r = 5.</li> <li>Simplify the equation: (x - 6) / (3 - 6) = -(y - 2) / (-2 - 2) = -(25 - (6 - 3)² - (2 + 2)²) / 5.

</section> <section style="font-size: 50px";> <h2 id="finding-intersection-points---example-7">Finding Intersection Points - Example 7</h2> <ul> <li>Find the intersection points of the functions f(x) = x^2 - 3 and g(x) = 2x - 5.</li> <li>Solution: <ul> <li>Set the equations equal to each other: x^2 - 3 = 2x - 5.</li> <li>Rearrange the equation: x^2 - 2x + 2 = 0.</li> <li>This is a quadratic equation, solve using factoring or quadratic formula.</li> <li>The solutions are x ≈ 1.732 and x ≈ 0.268.</li> <li>Substitute x = 1.732 into f(x) to find y ≈ -1.732.</li> <li>Substitute x = 0.268 into f(x) to find y ≈ -2.465.

</section> <section style="font-size: 50px";> <h2 id="example-8">Example 8</h2> <ul> <li>Find the intersection points of the functions f(x) = √x and g(x) = 4 - x.</li> <li>Solution: <ul> <li>Set the equations equal to each other: √x = 4 - x.</li> <li>Square both sides of the equation: x = 16 - 8x + x².</li> <li>Rearrange the equation: x² - 9x + 16 = 0.</li> <li>This is a quadratic equation, solve using factoring or quadratic formula.</li> <li>The solutions are x ≈ 1.631 and x ≈ 7.369.</li> <li>Substitute x = 1.631 into f(x) to find y ≈ 1.277.</li> <li>Substitute x = 7.369 into f(x) to find y ≈ -2.369.

</section> <section style="font-size: 50px";> <h2 id="summary-1">Summary</h2> <ul> <li>Equations of tangent and normals can be found using the derivative of a function.</li> <li>Intersection points can be found by solving the equations of the functions or lines.</li> <li>The equation of the normal to a circle can be found using the equation of the tangent line.</li> <li>Intersection points are useful in understanding the behavior of curves and lines.

</section> <section style="font-size: 50px";> <h2 id="example-9">Example 9</h2> <ul> <li>Consider the function f(x) = 3x^3 - 2x^2 + 5x - 1.</li> <li>Find the equation of the tangent line at the point (2, 23).</li> <li>Solution: <ul> <li>First, find the derivative of the function: f’(x) = 9x^2 - 4x + 5.</li> <li>Substitute x = 2 into the derivative to find f’(2) = 9(2)^2 - 4(2) + 5 = 29.</li> <li>The equation of the tangent line is: y - 23 = 29(x - 2).</li> <li>Simplify the equation: y = 29x - 37.

</section> <section style="font-size: 50px";> <h2 id="example-10">Example 10</h2> <ul> <li>Consider the function g(x) = e^x - x.</li> <li>Find the equation of the tangent line at the point (1, e-1).</li> <li>Solution: <ul> <li>First, find the derivative of the function: g’(x) = e^x - 1.</li> <li>Substitute x = 1 into the derivative to find g’(1) = e^1 - 1 = e - 1.</li> <li>The equation of the tangent line is: y - e^(-1) = (e - 1)(x - 1).</li> <li>Simplify the equation: y = (e - 1)x + 1 - e^(-1).

</section> <section style="font-size: 50px";> <h2 id="equation-of-normal-to-a-curve">Equation of Normal to a Curve</h2> <ul> <li> <p>The equation of the normal to a curve at a given point (x₁, y₁) can also be found using the derivative of the function.</p> </li> <li> <p>The equation of the normal line is given by: <code>y - y₁ = -1/f'(x₁)(x - x₁)</code></p> </li> <li> <p>Where:</p> <ul> <li>(x₁, y₁) is the given point on the curve</li> <li>f’(x₁) is the derivative of the function evaluated at x = x₁</li> </ul> </li> <li> <p>Notice that the slope of the normal line is the negative reciprocal of the slope of the tangent line.

</section> <section style="font-size: 50px";> <h2 id="example-11">Example 11</h2> <ul> <li>Consider the function h(x) = ln(x).</li> <li>Find the equation of the normal line at the point (2, ln 2).</li> <li>Solution: <ul> <li>First, find the derivative of the function: h’(x) = 1/x.</li> <li>Substitute x = 2 into the derivative to find h’(2) = 1/2.</li> <li>The equation of the normal line is: y - ln 2 = -(1/(1/2))(x - 2).</li> <li>Simplify the equation: y - ln 2 = -2(x - 2).

</section> <section style="font-size: 50px";> <h2 id="finding-intersection-points---example-12">Finding Intersection Points - Example 12</h2> <ul> <li>Find the intersection points of the functions f(x) = x^2 + 3x + 2 and g(x) = 2x - 1.</li> <li>Solution: <ul> <li>Set the equations equal to each other: x^2 + 3x + 2 = 2x - 1.</li> <li>Rearrange the equation: x^2 + x + 3 = 0.</li> <li>This is a quadratic equation, solve using factoring or quadratic formula.</li> <li>The solutions are x ≈ -0.682 and x ≈ -4.318.</li> <li>Substitute x = -0.682 into f(x) to find y ≈ 0.364.</li> <li>Substitute x = -4.318 into f(x) to find y ≈ 0.682.

</section> <section style="font-size: 50px";> <h2 id="example-13">Example 13</h2> <ul> <li>Find the intersection points of the functions f(x) = cos(x) and g(x) = sin(x).</li> <li>Solution: <ul> <li>Set the equations equal to each other: cos(x) = sin(x).</li> <li>Rearrange the equation: cos(x) - sin(x) = 0.</li> <li>This requires trigonometric identities to solve.</li> <li>The solutions are x = -π/4 and x = 5π/4.</li> <li>Substitute x = -π/4 into f(x) to find y = √2/2.</li> <li>Substitute x = 5π/4 into f(x) to find y = -√2/2.

</section> <section style="font-size: 50px";> <h2 id="summary-2">Summary</h2> <ul> <li>Equations of tangent and normals can be found using the derivative of a function.</li> <li>Intersection points can be found by solving the equations of the functions or lines.</li> <li>The equation of the normal to a curve can be found using the equation of the tangent line.</li> <li>Intersection points are useful in understanding the behavior of curves and lines.

</section> <section style="font-size: 50px";> <h2 id="summary-3">Summary</h2> <ul> <li>Equations of tangent and normals can be found using the derivative of a function.</li> <li>Intersection points can be found by solving the equations of the functions or lines.</li> <li>The equation of the normal to a curve can be found using the equation of the tangent line.</li> <li>Intersection points are useful in understanding the behavior of curves and lines.