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3 years ago

8

What is the solution set of y = x2 + 2x + 7 and y = x + 7? A. {(0, 7), (-1, 6)} B. {(0, 7), (-7, 0)} C. {(0, 7), (1, 8)} D. {(-2

, 0), (4, 0)}

1 answer:

Elena-2011 [213]3 years ago

8 0

Answer:

A.

Step-by-step explanation:

If

$y=x^2+2x+7$ AND

y = x + 7, then by the transitive property of equality:

$x^2+2x+7=x+7$

We can solve for the values of x by getting everything on one side of the equals sign and then solving for x:

$x^2+x=0$

We can factor out the common x to get:

x(x + 1) = 0

which tells us by the Zero Product Property that either

x = 0 and/or x + 1 = 0 and x = -1

We are expecting 2 solutions for x since this is a second degree polynomial. We will sub both -1 and 0 into y = x + 7 to solve for the corresponding values of y

y = 0 + 7 so

y = 7 and the coordinate is (0, 7)

y = -1 + 7 so

y = 6 and the coordinate is (-1, 6)

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Answer:

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2 years ago

In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En

mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

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When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

$\frac{dy}{dx} = 3t^{2} - 2t - 27$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

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Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

$x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55$

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The first point is (27.55, 7.22)

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