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question 11
let ( f(x) = ax^2 + bx + 4 ) and ( g(x) = 2x - 3 ). find ( a ) such that the graphs of ( f(x) ) and ( g(x) ) intersect when ( x = 2 ). if necessary, enter your answer as a decimal.
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Explanation:

Step1: Find f(2) and g(2)

When two graphs intersect at \( x = 2 \), their \( y \)-values are equal, so \( f(2)=g(2) \).

First, calculate \( g(2) \):
\( g(x)=2x - 3 \), substitute \( x = 2 \):
\( g(2)=2\times2 - 3=4 - 3 = 1 \)

Then, calculate \( f(2) \):
\( f(x)=Ax^{2}+bx + 4 \), substitute \( x = 2 \):
\( f(2)=A\times(2)^{2}+b\times2 + 4=4A + 2b + 4 \)

Wait, there might be a typo in the problem? Wait, the function \( g(x) \) is written as \( 2x - 3 \) or \( 2a - 3 \)? Wait, the original problem says \( g(x)=2x - 3 \) (probably a typo if it's \( 2a \), but assuming it's \( 2x \)). Wait, but let's check again. The problem says "Find \( A \) such that the graphs of \( f(x) \) and \( g(x) \) intersect when \( x = 2 \)". So at \( x = 2 \), \( f(2)=g(2) \).

Wait, maybe the \( g(x) \) is \( 2x - 3 \). Let's proceed.

So \( f(2)=g(2) \)

\( 4A + 2b + 4=1 \)

Wait, but we have two variables \( A \) and \( b \). Wait, maybe there's a mistake in the problem, or maybe \( b = 0 \)? Wait, no, maybe the original problem has \( g(x)=2x - 3 \) and maybe \( b \) is a typo, or maybe the problem is \( f(x)=Ax^{2}+4 \) (no \( bx \) term). Wait, maybe the user made a typo, but let's assume that maybe \( b = 0 \), or maybe the problem is \( f(x)=Ax^{2}+4 \) and \( g(x)=2x - 3 \). Let's check again.

Wait, the problem says "Let \( f(x)=Ax^{2}+bx + 4 \) and \( g(x)=2x - 3 \). Find \( A \) such that the graphs of \( f(x) \) and \( g(x) \) intersect when \( x = 2 \)."

But with two variables \( A \) and \( b \), we can't solve for \( A \) unless there's more information. Wait, maybe it's a typo and \( g(x)=2a - 3 \) is \( g(x)=2x - 3 \), and maybe \( b = 0 \), or maybe the problem is \( f(x)=Ax^{2}+4 \) (no \( bx \) term). Let's assume that \( b = 0 \) (maybe a typo). Then:

\( f(2)=A\times4 + 0 + 4=4A + 4 \)

\( g(2)=2\times2 - 3=1 \)

Set equal: \( 4A + 4 = 1 \)

\( 4A=1 - 4=-3 \)

\( A=\frac{-3}{4}=-0.75 \)

Wait, but maybe the \( g(x) \) is \( 2x - 3 \), and the \( f(x) \) has \( bx \) but maybe in the problem, \( b \) is a specific value, but since it's not given, maybe it's a mistake. Alternatively, maybe the original problem is \( f(x)=Ax^{2}+4 \) (no \( bx \) term). Let's proceed with that assumption.

Step1: Calculate g(2)

\( g(x)=2x - 3 \), so \( g(2)=2(2)-3 = 1 \)

Step2: Calculate f(2)

\( f(x)=Ax^{2}+4 \) (assuming \( bx = 0 \)), so \( f(2)=A(2)^{2}+4 = 4A + 4 \)

Step3: Set f(2)=g(2) and solve for A

\( 4A + 4 = 1 \)

Subtract 4 from both sides: \( 4A = 1 - 4 = -3 \)

Divide by 4: \( A=\frac{-3}{4}=-0.75 \)

Answer:

\( -0.75 \)