what is the equation of the line tangent to the graph of f(x)=7x-x^2 at the point where f(x)=3
6.4 Equation of a tangent to a curve (EMCH8)
At a given point on a curve, the slope of the curve is equal to the gradient of the tangent to the curve.
The derivative (or slope function) describes the gradient of a curve at whatsoever point on the curve. Similarly, it also describes the slope of a tangent to a curve at any point on the curve.
To determine the equation of a tangent to a curve:
- Find the derivative using the rules of differentiation.
- Substitute the \(x\)-coordinate of the given betoken into the derivative to calculate the gradient of the tangent.
- Substitute the gradient of the tangent and the coordinates of the given point into an appropriate form of the direct line equation.
- Brand \(y\) the subject area of the formula.
The normal to a curve is the line perpendicular to the tangent to the curve at a given point.
\[m_{\text{tangent}} \times m_{\text{normal}} = -ane\]Worked case thirteen: Finding the equation of a tangent to a curve
Find the equation of the tangent to the bend \(y=3{10}^{two}\) at the point \(\left(ane;iii\right)\). Sketch the curve and the tangent.
Find the derivative
Use the rules of differentiation:
\begin{align*} y &= 3{x}^{2} \\ & \\ \therefore \frac{dy}{dx} &= three \left( 2x \right) \\ &= 6x \stop{align*}
Calculate the gradient of the tangent
To determine the gradient of the tangent at the point \(\left(1;3\right)\), we substitute the \(x\)-value into the equation for the derivative.
\begin{align*} \frac{dy}{dx} &= 6x \\ \therefore grand &= 6(i) \\ &= half-dozen \terminate{align*}
Decide the equation of the tangent
Substitute the gradient of the tangent and the coordinates of the given point into the slope-signal form of the direct line equation.
\begin{marshal*} y-{y}_{1} & = thousand\left(x-{x}_{ane}\correct) \\ y-three & = half-dozen\left(x-1\right) \\ y & = 6x-6+3 \\ y & = 6x-3 \finish{marshal*}
Sketch the curve and the tangent
Worked instance 14: Finding the equation of a tangent to a curve
Given \(g(x)= (x + 2)(2x + ane)^{2}\), determine the equation of the tangent to the curve at \(10 = -one\) .
Determine the \(y\)-coordinate of the point
\begin{marshal*} g(x) &= (ten + 2)(2x + ane)^{2} \\ thousand(-1) &= (-1 + 2)[2(-i) + i]^{two} \\ &= (ane)(-1)^{2} \\ & = 1 \end{align*}
Therefore the tangent to the bend passes through the point \((-1;ane)\).
Expand and simplify the given function
\begin{align*} m(x) &= (x + 2)(2x + 1)^{ii} \\ &= (x + 2)(4x^{ii} + 4x + i) \\ &= 4x^{3} + 4x^{two} + x + 8x^{2} + 8x + 2 \\ &= 4x^{3} + 12x^{two} + 9x + 2 \end{marshal*}
Detect the derivative
\begin{align*} g'(10) &= 4(3x^{2}) + 12(2x) + ix + 0 \\ &= 12x^{ii} + 24x + 9 \end{align*}
Calculate the gradient of the tangent
Substitute \(x = -\text{1}\) into the equation for \(g'(x)\):
\begin{align*} k'(-1) &= 12(-1)^{2} + 24(-1) + 9 \\ \therefore 1000 &= 12 - 24 + nine \\ &= -3 \cease{align*}
Determine the equation of the tangent
Substitute the gradient of the tangent and the coordinates of the point into the gradient-point course of the directly line equation.
\begin{align*} y-{y}_{1} & = m\left(x-{x}_{one}\right) \\ y-1 & = -iii\left(ten-(-ane)\right) \\ y & = -3x - iii + one \\ y & = -3x - 2 \stop{align*}
Worked instance 15: Finding the equation of a normal to a curve
- Determine the equation of the normal to the bend \(xy = -4\) at \(\left(-1;iv\right)\).
- Draw a rough sketch.
Find the derivative
Make \(y\) the field of study of the formula and differentiate with respect to \(x\):
\begin{align*} y &= -\frac{4}{x} \\ &= -4x^{-ane} \\ & \\ \therefore \frac{dy}{dx} &= -4 \left( -1x^{-2} \correct) \\ &= 4x^{-two} \\ &= \frac{4}{x^{two}} \end{align*}
Calculate the gradient of the normal at \(\left(-one;4\right)\)
Offset make up one's mind the slope of the tangent at the given point:
\brainstorm{align*} \frac{dy}{dx} &= \frac{4}{(-i)^{2}} \\ \therefore one thousand &= 4 \end{align*}
Use the gradient of the tangent to summate the gradient of the normal:
\begin{align*} m_{\text{tangent}} \times m_{\text{normal}} &= -i \\ 4 \times m_{\text{normal}} &= -1 \\ \therefore m_{\text{normal}} &= -\frac{1}{four} \terminate{align*}
Find the equation of the normal
Substitute the gradient of the normal and the coordinates of the given bespeak into the gradient-point course of the straight line equation.
\begin{align*} y-{y}_{i} & = m\left(x-{ten}_{ane}\right) \\ y-4 & = -\frac{one}{4}\left(10-(-i)\right) \\ y & = -\frac{ane}{4}x - \frac{1}{4} + 4\\ y & = -\frac{1}{iv}x + \frac{15}{4} \end{align*}
Depict a rough sketch
Equation of a tangent to a bend
Textbook Exercise half-dozen.5
Determine the equation of the tangent to the bend defined by \(F(ten)=x^{iii}+2x^{2}-7x+ane\) at \(x=2\).
\begin{align*} \text{Gradient of tangent }&= F'(ten) \\ F'(x) &=3x^{2} +4x - vii \\ F'(2) &=3(2)^{2} + (4)(two) -7 \\ &=xiii \\ \therefore \text{Tangent: } y &=13x +c \cease{align*}
where \(c\) is the \(y\)-intercept.
Tangent meets \(F(x)\) at \((ii;F(two))\)
\brainstorm{align*} F(ii) &=(2)^{iii} + 2(ii)^{2} - 7(two) +i \\ &= 8 + 8 -14 +1 \\ &=3 \\ \text{Tangent: } three &=xiii(2) + c \\ \therefore c &= - 23 \\ y & = 13x - 23 \end{align*}
\(f(x)=i-3x^{two}\) is equal to \(\text{5}\).
\begin{align*} \text{Gradient of tangent } = f'(ten) = -6x \\ \therefore -6x &= 5 \\ \therefore x &= - \frac{5}{six} \\ \text{And } f\left(- \frac{5}{6} \right) &=1-3 \left( - \frac{5}{6} \right)^{two} \\ &=1-3 \left( \frac{25}{36} \right) \\ &=one - \frac{25}{12} \\ &= - \frac{thirteen}{12} \\ \therefore & \left( - \frac{five}{vi};- \frac{13}{12} \right) \terminate{align*}
\(g(x)=\frac{1}{3}x^{2}+2x+1\) is equal to \(\text{0}\).
\begin{align*} \text{Gradient of tangent } = yard'(x) = \frac{2}{3}10+ii \\ \therefore \frac{two}{3}x+2 &=0 \\ \frac{2}{3}x &= -two\\ \therefore x&=-2 \times \frac{3}{2} \\ &=-3 \\ \text{And } g(-three) &= \frac{1}{3}(-iii)^{2}+2(-3)+i \\ &= \frac{1}{3}(9)-6+1 \\ &= three-half dozen+one \\ &= -two \\ \therefore & (-3;-2) \terminate{align*}
parallel to the line \(y=4x-2\).
\begin{align*} \text{Slope of tangent }&= f'(x) \\ f(x)&=(2x-ane)^{two} \\ &= 4x^{2}-4x+1 \\ \therefore f'(x)&= 8x-four \\ \text{Tangent is parallel to } y&=4x-2 \\ \therefore m&=4 \\ \therefore f'(x) = 8x-4 &= iv \\ 8x &= 8 \\ x & = 1\\ \text{For } x=1: \quad y & = (ii(one)-1)^{two} \\ & = one \end{align*}
Therefore, the tangent is parallel to the given line at the point \((ane;1)\).
perpendicular to the line \(2y+ten-4=0\).
\begin{align*} \text{Perpendicular to } 2y + x - 4 &= 0 \\ y&= -\frac{ane}{ii}10+2\\ \therefore \text{ gradient of } \perp \text{ line } & = ii \quad (m_1 \times m_2 = -1) \\ \therefore f'(x) &= 8x-4 \\ \therefore 8x-4 &=2\\ 8x&=half-dozen\\ x&=\frac{three}{4} \\ \therefore y&=\left[2\left(\frac{3}{four}\right)-ane\right]^{2} \\ &=\frac{1}{4} \\ \therefore \left(\frac{3}{4};\frac{ane}{4}\right) \cease{align*}
Therefore, the tangent is perpendicular to the given line at the point \(\left(\frac{3}{iv};\frac{1}{iv}\correct)\).
Draw a graph of \(f\), indicating all intercepts and turning points.
Complete the square:
\begin{align*} y&=-[x^{two}-4x+three] \\ &=-[(ten-2)^{2}-4+3] \\ &=-(x-2)^{2}+ane\\ \text{Turning point}:&(ii;one) \end{align*} \(\text{Intercepts:}\\ y_{\text{int}}: x = 0, y = -iii \\ x_{\text{int}}: y=0, \\ -ten^{2} +4x -3 = 0 \\ ten^{2} - 4x + 3 = 0 \\ (10-3)(x-1) = 0 \\ x=three \text{ or } x=one \\ \text{Shape: "pout" } (a < 0) \\\) 
Find the equations of the tangents to \(f\) at:
- the \(y\)-intercept of \(f\).
- the turning bespeak of \(f\).
- the point where \(x = \text{four,25}\).
- \brainstorm{align*} y_{\text{int}}: (0;-3) \\ m_{\text{tangent}} = f'(x) &= -2x + four \\ f'(0) &=-2(0) + 4 \\ \therefore thousand &=4\\ \text{Tangent }y&=4x+c\\ \text{Through }(0;-3) \therefore y&=4x-3 \terminate{align*}
- \brainstorm{marshal*} \text{Turning bespeak: } (2;1) \\ m_{\text{tangent}} = f'(2) &= -2(2) + 4 \\ &=0\\ \text{Tangent equation } y &= ane \stop{align*}
- \begin{align*} \text{If } ten &=\text{4,25} \\ f(\text{4,25})&=-\text{iv,25}^{two}+4(\text{4,25})-3 \\ &= -\text{4,0625} \\ m_{\text{tangent}} \text{ at } x&= \text{4,25} \\ k&=-two(\text{4,25})+4\\ &=-\text{4,5} \\ \text{Tangent }y&=-\text{4,v}10+c\\ \text{Through }(\text{4,25};-\text{four,0625}) \\ -\text{4,0625}&=-\text{four,five}(\text{4,25})+c\\ \therefore c&= \text{fifteen,0625} \\ y&=-\text{4,v}x+\text{xv,0625} \end{align*}
Draw the three tangents above on your graph of \(f\).
Write downwards all observations about the three tangents to \(f\).
Tangent at \(y_{\text{int}}\) (bluish line): gradient is positive, the function is increasing at this signal.
Tangent at turning point (light-green line): slope is zero, tangent is a horizontal line, parallel to \(10\)-axis.
Tangent at \(10=\text{4,25}\) (regal line): gradient is negative, the function is decreasing at this betoken.
Source: https://intl.siyavula.com/read/maths/grade-12/differential-calculus/06-differential-calculus-04
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