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mina [271]
1 year ago
12

Find features such as x and y intercepts to sketch this function (using e or in is not allowed)

Mathematics
1 answer:
Deffense [45]1 year ago
7 0

Let's find the x-intercept of the given function. This ocurrs when y is equal to zero, so we have

0=-log_2(3(x+2))+4

which gives

4-log_2(3(x+2))=0

Now, we can rewrite the number 4 as follows

4=log_22^4

So, by substituting this result into the above equation, we have

log_22^4-log_2(3(x+2))=0

From the quotient rule of the logarithms, it can be written as

log_2\frac{2^4}{3(x+2)}=0

From the property

log_b1=0

we can conclude that

\frac{2^4}{3(x+2)}=1

or equivalently,

\frac{16}{3(x+2)}=1

so, we have

3(x+2)=16

which gives

\begin{gathered} x+2=\frac{16}{3} \\ then \\ x=\frac{16}{3}+2=7.3333 \end{gathered}

So, we have obtained that the x-intercept is the point (7.333, 0).

Similarly, the y-intercept ocurrs at x=0, which implies that

y=-log_2(3(0+2))+4

or equivalently,

y=-log_2(-6)+4

However, for a real base (2 in our case) the logarithm is undefined. This fact and since the logarithm has negative coefficient mean that the graph of the function has the form:

As we can corroborate with the followin graph:

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Answer:

(a) t = 7 sec approximately; (b) t = 6 sec

Step-by-step explanation:

(a)  Set h(t)= -16(t-3)^2 + 288 = 0 and solve for t:

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       After simplification, this becomes (t - 3)^2 = 18, or t - 3 = ±3√2.

        Because t can be only zero or positive, t = 3 + 3√2 = 7 seconds

(b) Solve h(t)= -16(t-3)^2 + 288 = 150:

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