The functions f(x)=-(x-4)^(2)+3 and g(x)=-(x-4)^(2)+5 are graphed in the same xy-plane as y=f(x) and y=g(x). If (h_(1),k_(1)) is the vertex of the graph of function f and (h_(2),k_(2)) is the vertex of the graph of function g, then what is k_(2)-k_(1) ? Choose 1 answer: (A) -2 (B) 0 (c) 2 (D) 8

Identifying the vertex of f(x): The question prompt is: ＂What is the difference in the y-coordinates of the vertices of the graphs of the functions f(x) and g(x)?＂Identifying the vertex of g(x): First, let's identify the vertex of the graph of function f(x). The function f(x) is in the form of a parabola, f(x) = a(x-h)^2 + k, where (h, k) is the vertex of the parabola. For f(x) = -(x-4)^2 + 3, the vertex (h_1, k_1) is (4, 3).Calculating the difference in y-coordinates: Next, let's identify the vertex of the graph of function g(x). The function g(x) is also in the form of a parabola, g(x) = a(x-h)^2 + k. For g(x) = -(x-4)^2 + 5, the vertex (h_2, k_2) is (4, 5).Now, we need to find the difference in the y-coordinates of the vertices of the two functions. This is calculated by subtracting k_1 from k_2: k_2 - k_1 = 5 - 3.Performing the subtraction, we get k_2 - k_1 = 5 - 3 = 2.

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The functions
f(x)=-(x-4)^(2)+3 and
g(x)=-(x-4)^(2)+5 are graphed in the same
xy-plane as
y=f(x) and
y=g(x). If
(h_(1),k_(1)) is the vertex of the graph of function
f and
(h_(2),k_(2)) is the vertex of the graph of function
g, then what is
k_(2)-k_(1) ?
Choose 1 answer:
(A) -2
(B) 0
(c) 2
(D) 8

The functions $f(x)=-(x-4)^{2}+3$ and $g(x)=-(x-4)^{2}+5$ are graphed in the same $x y$ -plane as $y=f(x)$ and $y=g(x)$ . If $\left(h_{1}, k_{1}\right)$ is the vertex of the graph of function $f$ and $\left(h_{2}, k_{2}\right)$ is the vertex of the graph of function $g$ , then what is $k_{2}-k_{1}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-2$ $\newline$ (B) $0$ $\newline$ (C) $2$ $\newline$ (D) $8$

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Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

Q. The functions

f(x)=-(x-4)^{2}+3

and

g(x)=-(x-4)^{2}+5

are graphed in the same

x y

-plane as

y=f(x)

and

y=g(x)

. If

\left(h_{1}, k_{1}\right)

is the vertex of the graph of function

f

and

\left(h_{2}, k_{2}\right)

is the vertex of the graph of function

g

, then what is

k_{2}-k_{1}

\newline

Choose

1

answer:

\newline

(A)

-2

\newline

(B)

0

\newline

(C)

2

\newline

(D)

8

Identifying the vertex of $f(x)$ : The question prompt is: ＂What is the difference in the $y$ -coordinates of the vertices of the graphs of the functions $f(x)$ and $g(x)$ ?＂
Identifying the vertex of g(x): First, let's identify the vertex of the graph of function f(x). The function f(x) is in the form of a parabola, $f(x) = a(x-h)^2 + k$ , where $(h, k)$ is the vertex of the parabola. For $f(x) = -(x-4)^2 + 3$ , the vertex $(h_1, k_1)$ is $(4, 3)$ .
Calculating the difference in y-coordinates: Next, let's identify the vertex of the graph of function $g(x)$ . The function $g(x)$ is also in the form of a parabola, $g(x) = a(x-h)^2 + k$ . For $g(x) = -(x-4)^2 + 5$ , the vertex $(h_2, k_2)$ is $(4, 5)$ .
Calculating the difference in y-coordinates: Next, let's identify the vertex of the graph of function $g(x)$ . The function $g(x)$ is also in the form of a parabola, $g(x) = a(x-h)^2 + k$ . For $g(x) = -(x-4)^2 + 5$ , the vertex $(h_2, k_2)$ is $(4, 5)$ .Now, we need to find the difference in the y-coordinates of the vertices of the two functions. This is calculated by subtracting $k_1$ from $k_2$ : $k_2 - k_1 = 5 - 3$ .
Calculating the difference in y-coordinates: Next, let's identify the vertex of the graph of function $g(x)$ . The function $g(x)$ is also in the form of a parabola, $g(x) = a(x-h)^2 + k$ . For $g(x) = -(x-4)^2 + 5$ , the vertex $(h_2, k_2)$ is $(4, 5)$ .Now, we need to find the difference in the y-coordinates of the vertices of the two functions. This is calculated by subtracting $k_1$ from $k_2$ : $k_2 - k_1 = 5 - 3$ .Performing the subtraction, we get $k_2 - k_1 = 5 - 3 = 2$ .