#12 I asked this question already, but the person explained it a way that I haven’t been taught so I didn’t understand it at all… I’ve attached my work so you can see how we are being told to work it. We use the difference quotient equation. The image contains handwritten mathematical calculations that appear to be focused on determining a derivative and solving equations for a quadratic function. Below is an organized transcription suitable for an educational website.---### Calculations and Explanations #### Given Function\[ y = d \cdot ax - x^2 \]#### Points for CalculationPoints used: $ (3, 5.7) $#### Deriving Equation of a Line 1. **Calculate the slope** between points using: \[ y - 5.7 = -2.9(x - 3) \] \[ y - 5.7 = -2ax + 8.7 \] Simplifying: \[ y = -2x + 8.7 \]#### Derivative Using First Principles1. **Setting up the difference quotient:** \[ f'(x) = \lim_{{h \to 0}} \frac{4.9(x+h)^2 - (4.9(x) - x^2)}{h} \]2. **Simplifying the equation:** \[ 4.9(x+h)^2 - x^2 - 8xh - h^2 \] \[ \frac{4.9(x+h)^2 - (x)^2}{h} \]3. **Simplified Derivative Expression:** \[ 4.9x + 4.9h - x^2 - 8xh - h^2 \] Alongside the differentiation in the numerator, substitute to find immediate values for derivative test: \[ d + 4.9 - 2x - h^2 \]4. **Final evaluation for specific value:** \[ -h + (4.9 - 2x) \] Substituting specific numbers: \[ (0) + 4.9 - 2(3) = 4.9 - 6 = -2.9 \]---### Notes:- The given function is quadratic in nature.- Calculations involve both finding an equation of the tangent line and using the definition of the derivative (the limit of the difference quotient as $ h \to 0 $) for verification of slope at specific points.- The slope is evaluated at $ **Question:** Find an equation of the tangent line to the curve at the given point.Curve: \( y = 4.9x - x^2 $ Point: $ (3, 5.7) $**Attempted Solution:**Equation: $ y = 5.7x + 11.4 $ *This attempt is marked incorrect with a red 'X' symbol.***Additional Materials** section is shown below the problem. **Explanation:** The task involves finding the tangent line's equation at a specified point on the curve. The given incorrect solution suggests reevaluating the derivative to find the correct slope of the tangent, then using point-slope form to derive the correct equation.

Equation of a tangent line at a point (x0,y0) on the curve y=y(x) is y-y0=y'(x0)x-x0

Answered: Find an equation of the tangent line to…

Math

Calculus

Find an equation of the tangent line to the curve at the given point. y = 4.9x – x², (3, 5.7) - y= 5.7x + 11.4

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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