QuestionGiven the function $g(x)=2 x+6$ . $x$ -interept $(y=0)$ The function of $f(x)=b^{x}+9=2 x+6$ $0<b<1$ $\quad \begin{array}{ll}\frac{6}{2}= & 0=2 \quad(3 ; 0)\end{array}$ - $\quad 0<b<1$ - $f(0)=3$ $\begin{array}{ll} 0=2 x+6 & y \\ \frac{6}{2}=x(3 ; 0) & =6(0 ; 6) \end{array}$ - $f(0)=3$ range is $y \in \square, y>2$ 7.1 Draw a neat sketch of $f$ and $g$ on same set of axes. Clearly show all intercepts (5) 7.2 The coordinates of the point of intersection of $f$ and $g$ are $(-1 ; 4)$ . 7.2 .1
If $f$ and $g$ are reflected along the line $x=1$ , write down the coordinates of the point of intersection of the images of $f$ and $g$ under this transformation. 7.2.2 Determine the values of $x$ for which $0 \leq g(x) \leq 4$ 7.2.3 Determine the values of $k$ for which the roots of $-2 . f(x)=k$ are negative.

Studdy Solution
Determine values of $k$ for which the roots of $-2 \cdot f(x) = k$ are negative:
1. Set $-2(b^x + 9) = k$ .
2. Solve for $x$ when the roots are negative.
Since $b^x$ is an exponential function and $0 < b < 1$ , $b^x$ decreases as $x$ increases.
The inequality $-2(b^x + 9) = k$ implies:
$-2b^x - 18 = k$
For roots to be negative, solve for $x$ such that:
$b^x < -\frac{k + 18}{2}$
Since $b^x$ is always positive, $k + 18$ must be negative, implying $k < -18$ .
The solution for each part is:
1. The $x$ -intercept of $g(x)$ is $(-3, 0)$ .
2. The reflection of the intersection point is $(3, 4)$ .
3. The values of $x$ for $0 \leq g(x) \leq 4$ are $[-3, -1]$ .
4. The values of $k$ for which the roots of $-2 \cdot f(x) = k$ are negative are $k < -18$ .

View Full Solution - Free

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms