Problem 47
Question
\(47-50\) The graph of a polynomial function is given. From the graph, find (a) the \(x-\) and \(y\) -intercepts, and (b) the coordinates of all local extrema. $$ P(x)=-x^{2}+4 x $$
Step-by-Step Solution
Verified Answer
x-intercepts are (0,0) and (4,0); y-intercept is (0,0); local maximum at (2,4).
1Step 1: Identify the x-intercepts
To find the x-intercepts of the polynomial function, we need to solve for \(x\) when \(P(x) = 0\). That is, solve the equation \(-x^2 + 4x = 0\). Factoring this equation, we get \(-x(x - 4) = 0\), which gives the solutions \(x = 0\) and \(x = 4\). Therefore, the x-intercepts are at \((0, 0)\) and \((4, 0)\).
2Step 2: Identify the y-intercept
The y-intercept occurs where the graph crosses the y-axis, which is at \(x = 0\). To find this, substitute \(x = 0\) into the function: \(P(0) = -0^2 + 4(0) = 0\). Hence, the y-intercept is at \((0, 0)\).
3Step 3: Find the derivative for extrema
To find the local extrema, take the derivative of \(P(x)\). The derivative \(P'(x) = \frac{d}{dx}(-x^2 + 4x) = -2x + 4\).
4Step 4: Solve for critical points
Critical points occur where the first derivative is zero or undefined. Set the derivative equal to zero: \(-2x + 4 = 0\). Solving for \(x\), we get \(x = 2\).
5Step 5: Determine the nature of the critical point
To determine if the critical point at \(x = 2\) is a local maximum or minimum, use the second derivative test. The second derivative \(P''(x) = -2\). Since \(P''(x) < 0\), the function is concave down, indicating a local maximum at \(x = 2\).
6Step 6: Find the coordinates of the local maximum
To find the coordinates of the local maximum, substitute \(x = 2\) back into the original function: \(P(2) = -2^2 + 4(2) = -4 + 8 = 4\). Therefore, the local maximum is at \((2, 4)\).
Key Concepts
X-Intercepts in Polynomial FunctionsY-Intercepts in Polynomial FunctionsLocal Extrema in Polynomial Functions
X-Intercepts in Polynomial Functions
When identifying the x-intercepts of a polynomial function like \(-x^2 + 4x\), we are looking for the points where the graph crosses the x-axis. These occur at values of \(x\) for which the function \(P(x) = 0\).
In simpler terms, we solve the equation for zero.For the given polynomial, we set \(-x^2 + 4x = 0\). By factoring, we can rewrite this as \(-x(x - 4) = 0\). This equation tells us that if either part equals zero, the whole equation does:
Think of these intercepts as the ground where a projectile lands and takes off on the curve, portraying the zeros of the polynomial.
In simpler terms, we solve the equation for zero.For the given polynomial, we set \(-x^2 + 4x = 0\). By factoring, we can rewrite this as \(-x(x - 4) = 0\). This equation tells us that if either part equals zero, the whole equation does:
- \(x = 0\)
- \(x = 4\)
Think of these intercepts as the ground where a projectile lands and takes off on the curve, portraying the zeros of the polynomial.
Y-Intercepts in Polynomial Functions
The y-intercept of a polynomial function is found at the point where the graph crosses the y-axis. This happens where \(x = 0\), so to find it, you simply substitute 0 for \(x\) in the function.
In the function \(P(x) = -x^2 + 4x\), substituting \(x = 0\) means calculating \(P(0)\):
\[P(0) = -(0)^2 + 4(0) = 0\] Therefore, the y-intercept is at the point \((0, 0)\).
The y-intercept often provides a simple check-point on the graph, confirming how the function behaves at the vertical axis.It's like the starting height if we're imagining the graph as a jump or drop of some kind.
In the function \(P(x) = -x^2 + 4x\), substituting \(x = 0\) means calculating \(P(0)\):
\[P(0) = -(0)^2 + 4(0) = 0\] Therefore, the y-intercept is at the point \((0, 0)\).
The y-intercept often provides a simple check-point on the graph, confirming how the function behaves at the vertical axis.It's like the starting height if we're imagining the graph as a jump or drop of some kind.
Local Extrema in Polynomial Functions
Local extrema of a polynomial function refer to the highest or lowest points in a particular region of the graph. To find these points in \(-x^2 + 4x\), take the function's derivative to study its slope behavior. This helps in identifying critical points.
Given that the derivative is \(P'(x) = -2x + 4\), we find critical points by setting \(P'(x) = 0\):
\[-2x + 4 = 0\] Solving this, we get \(x = 2\). This critical point needs to be tested further to decide if it's a maximum or minimum.
By evaluating the second derivative \(P''(x) = -2\), we notice it's negative, implying the function is concave down. Therefore, there's a local maximum at \(x = 2\).
Substituting \(x = 2\) back into the original function gives the coordinates of the local maximum:
\[P(2) = -2^2 + 4(2) = 4\] Thus, the local maximum is at \((2, 4)\).
The local extremum highlights significant points of tension or relaxation, much like peaks or valleys along a mountainous journey.
Given that the derivative is \(P'(x) = -2x + 4\), we find critical points by setting \(P'(x) = 0\):
\[-2x + 4 = 0\] Solving this, we get \(x = 2\). This critical point needs to be tested further to decide if it's a maximum or minimum.
By evaluating the second derivative \(P''(x) = -2\), we notice it's negative, implying the function is concave down. Therefore, there's a local maximum at \(x = 2\).
Substituting \(x = 2\) back into the original function gives the coordinates of the local maximum:
\[P(2) = -2^2 + 4(2) = 4\] Thus, the local maximum is at \((2, 4)\).
The local extremum highlights significant points of tension or relaxation, much like peaks or valleys along a mountainous journey.
Other exercises in this chapter
Problem 47
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