Consider the function:(-2x^2 +3x -7) y= -2x^2x +3x +17

Find y ′ using implicit differentiation. Do not solve for y.

What is the slope of the tangent at (x,y) = (−1,−1)?

Find y ′ by solving for y and using the quotient rule.

What is the slope of the tangent when x = -1 ?

Thanks for your help.

(-2x^2 +3x -7) y= -2x^2 +3x +17

(-4x+3)y + (-2x^2+3x-7)y' = -4x+3

y' = (4x-3)(y-1)/(2x^2-3x+7)
y'(-1,-1) = (-7)(-2)/(2+3+7) = 7/6

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doing the long division,

y = (2x^2-3x-17)/(2x^2-3x+7)
= 1 - 24/(2x^2-3x+7)

y' = 24(4x-3)/(2x^2-3x-7)^2

y'(-1) = 24(-7)/12^2 = -7/6
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using the quotient rule,

y = (2x^2-3x-17)/(2x^2-3x+7)
y' = [(4x-3)(2x^2-3x+7)-(2x^2-3x-17)(4x-3)]/(2x^2-3x+7)^2
= 24(4x-3)/(2x^2-3x+7)^2

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The tangent line is thus y+1 = -7/6 (x+1)

Check the graphs at

http://www.wolframalpha.com/input/?i=plot+(2x%5E2-3x-17)%2F(2x%5E2-3x%2B7),+y%3D(-7%2F6)(x%2B1)-1

To find the derivative of y with respect to x, we need to use implicit differentiation.

1. To find y', differentiate each term with respect to x.
-2x^2 + 3x - 7 will become -4x + 3.
-2x^2x + 3x + 17 will become -2x^2 + 1.

So, we have the equation -4x + 3 = -2x^2 + 1.

2. To find the slope of the tangent at (x,y) = (-1,-1), substitute x = -1 into the equation -4x + 3 = -2x^2 + 1 and solve for y.
-4(-1) + 3 = -2(-1)^2 + 1
4 + 3 = -2 + 1
7 = -1

Therefore, the slope of the tangent at (x,y) = (-1,-1) is -1.

3. To find y' by solving for y and using the quotient rule, we need to rearrange the equation -4x + 3 = -2x^2 + 1 to solve for y.
Start by moving all the terms to one side.
-2x^2 + 4x - 2 = 0

Now, we have a quadratic equation -2x^2 + 4x - 2 = 0. Solve this quadratic equation for x.

Let's use the quadratic formula: x = (-b ± √(b^2 - 4ac))/(2a)

Plugging in a = -2, b = 4, c = -2, we get:
x = (-4 ± √(4^2 - 4*(-2)*(-2)))/(2*(-2))
x = (-4 ± √(16 - 16))/(-4)
x = (-4 ± √0)/(-4)
x = -1

So, we have x = -1 as the solution for the quadratic equation -2x^2 + 4x - 2 = 0.

4. Once we have the solution for x, substitute it back into the original equation -2x^2 + 1 to find y.
With x = -1, we have y = -2(-1)^2 + 1
y = -2(1) + 1
y = -2 + 1
y = -1

Therefore, the value of y when x = -1 is -1.

5. Finally, to find the slope of the tangent when x = -1, we need to find y'.
We previously found that y = -1 when x = -1, so substitute these values into the equation -2x^2 + 1 and solve for y'.
y' = -2(-1)^2 + 1
y' = -2(1) + 1
y' = -2 + 1
y' = -1

Therefore, the slope of the tangent when x = -1 is -1.