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FinnZ [79.3K]
2 years ago
15

A line passes through the point (-4,-5) and had a slope of 5/2. Write an equation in slope-intercept form

Mathematics
1 answer:
daser333 [38]2 years ago
4 0

Step-by-step explanation:

as we have a point and the slope, we can start with the point-slope form and then transform.

the point-slope form is

y - y1 = a(x - x1)

(x1, y1) being a point on the line, a being the slope.

the slope-interceot form is

y = ax + b

a being the slope again, b being the y-intercept (the y value for x = 0).

so, we have

y - -5 = 5/2 × (x - -4)

y + 5 = 5/2 × (x + 4) = 5x/2 + 5×4/2 = 5x/2 + 10

y = 5x/2 + 5

or

y = (5/2)x + 5

and this is already the slope-intercept form. all done.

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F(x) = e^-x . Find the equation of the tangent to f(x) at x=-1​
natima [27]

Answer:

The <em>equation</em> of the tangent line is given by the following equation:

\displaystyle y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg)

General Formulas and Concepts:

<u>Algebra I</u>

Point-Slope Form: y - y₁ = m(x - x₁)

  • x₁ - x coordinate
  • y₁ - y coordinate
  • m - slope

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

*Note:

Recall that the definition of the derivative is the <em>slope of the tangent line</em>.

<u>Step 1: Define</u>

<em>Identify given.</em>

<em />\displaystylef(x) = e^{-x} \\x = -1

<u>Step 2: Differentiate</u>

  1. [Function] Apply Exponential Differentiation [Derivative Rule - Chain Rule]:
    \displaystyle f'(x) = e^{-x}(-x)'
  2. [Derivative] Rewrite [Derivative Rule - Multiplied Constant]:
    \displaystyle f'(x) = -e^{-x}(x)'
  3. [Derivative] Apply Derivative Rule [Derivative Rule - Basic Power Rule]:
    \displaystyle f'(x) = -e^{-x}

<u>Step 3: Find Tangent Slope</u>

  1. [Derivative] Substitute in <em>x</em> = 1:
    \displaystyle f'(1) = -e^{-1}
  2. Rewrite:
    \displaystyle f'(1) = \frac{-1}{e}

∴ the slope of the tangent line is equal to  \displaystyle \frac{-1}{e}.

<u>Step 4: Find Equation</u>

  1. [Function] Substitute in <em>x</em> = 1:
    \displaystyle f(1) = e^{-1}
  2. Rewrite:
    \displaystyle f(1) = \frac{1}{e}

∴ our point is equal to  \displaystyle \bigg( 1, \frac{1}{e} \bigg).

Substituting in our variables we found into the point-slope form general equation, we get our final answer of:

\displaystyle \boxed{ y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg) }

∴ we have our final answer.

---

Learn more about derivatives: brainly.com/question/27163229

Learn more about calculus: brainly.com/question/23558817

---

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

3 0
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