$x^2+13x=-42\ \ \ |+42\\\\x^2+13x+42=0\\\\x^2+6x+7x+42=0\\\\x(x+6)+7(x+6)=0\\\\(x+6)(x+7)=0\iff x+6=0\ \vee\ x+7=0\\\\x=-6\ \vee\ x=-7$

Anaswer: a. x = -6 or x = -7

8 0

3 years ago

Can you help me create a polynomial function matching the criteria on the left please and thanks

Olenka [21]

Answer:

$f(x) = \dfrac{1}{8}(x + 4)^2(x + 1)(x - 3)$

Step-by-step explanation:

From inspection of the graph, we can see that the curve intercepts the x-axis at (-4, 0), (-1, 0) and (3, 0)

Therefore,

x = -4 ⇒ x + 4 = 0

x = -1 ⇒ x + 1 = 0

x = 3 ⇒ x - 3 = 0

Because (-4, 0) touches the x-axis, then (x + 4)² will be a factor

So (x + 4)², (x + 1) and (x - 3) are all factors of the polynomial

$\implies f(x) = \dfrac{1}{n}(x + 4)^2(x + 1)(x - 3)$

If we multiply the constants, this will give us the y-intercept:

⇒ 4² x 1 x -3 = -48

From inspection of the graph, the y-intercept is -6

So to get from -48 to -6 we need to multiply -48 by 1/8

Therefore, n = 1/8

$\implies f(x) = \dfrac{1}{8}(x + 4)^2(x + 1)(x - 3)$

7 0

2 years ago

Read 2 more answers

Find the 15th term of the arithmetic sequence 13,8,3,-2

Lunna [17]

Answer:

-57

Step-by-step explanation:

13,8,3,-2 (decreasing by 5)

13,8,3,-2,-7,-12,-17,-22,-27,-32,-37,-42,-47,-52,-57.

The 15th term is -57.

Ace

4 0

2 years ago