ASAP! GIVING BRAINLIEST! Please read the question THEN answer correctly! No guessing.

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

7 0

Answer:

B. The graph flips over the x-axis

Step-by-step explanation:

Properties of : c · f(x)

- Graph opens wider/more narrow

- If the absolute value of c is greater than one, the shape narrows

- If the absolute value of c is less than one, it widens

- It can also affect any applied horizontal/vertical shifts

- If c is negative, the shape is reflected over the x-axis (upside-down)

In this case, the question is only asking if the function flips a certain way. As seen, "c," (as shown above) is -1, which is less than one. This means that the function will be flipped over the x-axis.

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<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$