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3 years ago

Solve it I need help

Mathematics

1 answer:

rusak2 [61]3 years ago

6 0

It opens upward and the vertex is (2,-6) here are some plotted points: (0,6),(1,-3),(2,-6),(3,-3),(4,6). Hope that helps. I hate graphing but if you do it enough times you'll get the hang of it!

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What is 2/3 of $300?

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Patty needs a total of $80 to buy a bicycle. She has already saved $35. If she saves $10 a week from her earnings, what is the l

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Martin finds a shirt on sale for 20% off at a department store. The original price was $20. Martin must also pay a 7.5% sales ta

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Answer: You will pay $13.52 for this shirt.

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What value of x will make ONM similar SRQ by the SAS similarity theorem?

Marta_Voda [28]

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See the attached figure.

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alexira [117]

$f(x) = x^3 - x - 1$

To find the gradient of the tangent, we must first differentiate the function.

$f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1$

The gradient at x = 0 is given by evaluating f'(0).

$f'(0) = 3(0)^2 - 1 = -1$

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

$y = -x + c$

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

$f(0) = (0)^3 - (0) - 1 = -1$

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

$y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}$

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