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3 years ago

Whats the table for y=-4|x-5|+3 how to solve absolute value function with a table and graph

Mathematics

1 answer:

zepelin [54]3 years ago

3 0

Answer:

For a table for x and y values for this absolute value equation, it would look like this:

x --- y

3 --- (-5)

4 --- (-1)

5 --- 3

6 --- (-1)

7 --- (-5)

Step-by-step explanation:

When you are building a table for an absolute value graph, you start with the base formula for absolute value equations:

y = a|x-h| + k

In this equation (h, k) is the vertex and therefore the middle point. From there we go two numbers in each direction for our x values. And for every change in x, y changes by a factor of a.

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Simplify<br><br> √12m • √15m<br><br> = [1] m √[2]

Elodia [21]

Answer:

6m√5

Step-by-step explanation:

√(12m)·√(15m) = √(180m²) = √(5·(36m²)) = √5·√(6m)²

= 6m√5 . . . . . . for m ≥ 0

8 0

3 years ago

What is the product of (7x+3)^2

Natali [406]

Answer:

49x² + 42x + 9

Step-by-step explanation:

Given

(7x + 3)²

= (7x + 3)(7x + 3)

Each term in the second factor is multiplied by each term in the first factor, that is

7x(7x + 3) + 3(7x + 3) ← distribute both parenthesis

= 49x² + 21x + 21x + 9 ← collect like terms

= 49x² + 42x + 9

7 0

2 years ago

Read 2 more answers

#78 please explain work

kenny6666 [7]

Answer:

$0.40c + $0.20h = $11

Step-by-step explanation:

c = chocolate pieces

h = hard candy

$0.40c + $0.20h = $11

5 0

3 years ago

6000 increased by 145%

Lisa [10]

Your question would be clearer if written as "What is 145% of 6000?"

1.45(6000) = 8700

8 0

3 years ago

f(x) is a quadratic function with x-intercepts at (−1, 0) and (−3, 0). If the range of f(x) is [−4, [infinity]) and g(x) = 2x^2

iren2701 [21]

Answer with Step-by-step explanation:

We are given that function f(x) which is quadratic function.

x -intercept of function f(x) at (-1,0) and (-3,0)

x-Intercept of f means zeroes of f

x=-1 and x=-3

Range of f =[-4, $\infty$ )

g(x)= $2x^2+8x+6=2(x^2+4x+3)$

$g(x)=0$

$2(x^2+4x+3)=0$

$x^2+4x+3=0$

$x^2+3x+x+3=0$

$x(x+3)+1(x+3)=0$

$(x+1)(x+3)=0$

$x+1=0\implies x=-1$

$x+3=0\implies x=-3$

Therefore, x-intercept of g(x) at (-1,0) and (-3,0).

Substitute x=-2

$g(-2)=2(-2)^2+8(-2)+6=8-16+6=-2$

$g(x)=2(x^2+4x)+6$

$g(x)=2(x^2+2\times x\times 2+4-4)+6=2(x^2+2\times x\times 2+4)-8+6$

$g(x)=2(x+2)^2-2$

By comparing with the equation of parabola

$y=a(x-h)^2+k$

Where vertex=(h,k)

We get vertex of g(x)=(-2,-2)

Range of g(x)=[-2, $\infty$ )

Zeroes of f and g are same .

But range of f and g are different.

Range of f contains -3 and -4 but range of g does not contain -3 and -4.

f and g are both quadratic functions.

3 0

3 years ago