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Andrej [43]
2 years ago
13

Find the equations of the tangents to y=x^2 which pass through the point (2,3).

Mathematics
1 answer:
algol [13]2 years ago
6 0

Answer: The equations of the tangents that pass through

y  =  -x- 1 and

y = 11x -25

Step-by-step explanation: The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So for our curve (the parabola) we have

y= x^2 +x

Differentiating wrt  x we get:

dy/dx = 2x +1

Let P (a, B)  be any generic point on the curve. Then the gradient of the tangent at P is given by:

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Simplify the expression -(2m+3)+5m
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Step-by-step explanation:

When equations are equivalent, they will have infinitely many solutions.

If you want to find an equivalent equation, you can <u>multiply/divide every</u> term in the equation by the same number. You can also <u>rearrange</u> the equation.

Since the options mostly have "y" by itself, start by <u>rearranging</u> to isolate "y" in the given equation.

Move all the other numbers to the right of the equal sign except for "y". To move a number, apply its reverse operation to the entire equation.

Do reverse operations in reverse BEDMAS order.

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y = 2x + 3

This equation looks like option 3 except every term has a different negative/positive.

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We can <u>multiply every term</u> by -1.

y = 2x + 3

y*-1 = 2x*-1 + 3*-1

-y = -2x + (-3)                Negative and positive make a negative

-y = -2x - 3                     Same as option 3

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