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2 years ago

What is the vertex of the quadratic function below?

1 answer:

3 0

$y = - 3 {x}^{2} - 12x + 17 \\ x_0 = \frac{ - b}{2a } = \frac{12}{ - 6} = - 2 \\ y_0 = - 3 \times {( - 2)}^{2} - 12 \times ( - 2) + 17 = 29$

Maybe i did something wrong, maybe your question is incorrect, but the vertex is (-2; 29)

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Find the solution set of this inequality|10x+20| ≤10

Pavlova-9 [17]

Answer:

solution is

[-3,-1]

Step-by-step explanation:

we are given

$|10x+20|\leq 10$

Firstly, we will find critical values

so, let's assume it is equal

$|10x+20|= 10$

now, we can break absolute sign

For $|10x+20|= -(10x+20)$ :

$-(10x+20)= 10$

we can solve for x

$-10x-20= 10$

Add both sides by 20

$-10x-20+20= 10+20$

$-10x= 30$

Divide both sides by -10

and we get

$x=-3$

For $|10x+20|= (10x+20)$ :

$(10x+20)= 10$

we can solve for x

$10x+20= 10$

Subtract both sides by 20

$10x+20-20= 10-20$

$10x= -10$

Divide both sides by 10

and we get

$x=-1$

so, critical values are

$x=-3$

$x=-1$

now, we can draw a number line and locate these values

and then we can check inequality on each intervals

For $(-\infty,-3)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-5

$|10\times -5+20|\leq 10$

$|-50+20|\leq 10$

$30\leq 10$

so, this is FALSE

For $[-3,-1]$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-2

$|10\times -2+20|\leq 10$

$|-20+20|\leq 10$

$0\leq 10$

so, this is TRUE

For $(-1,\infty)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=0

$|10\times 0+20|\leq 10$

$|0+20|\leq 10$

$20\leq 10$

so, this is FALSE

so, solution is

[-3,-1]

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What is the answer to this question??

Tasya [4]

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Artyom0805 [142]

so I think 37\12 is the answer to this question

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3 years ago

Read 2 more answers

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As part of a science experiment, Sam recorded the four values shown in the table, but one of them got smudged on his paper. He r

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3 years ago