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Question
given the function $c(x) = -(x - 5)(x + 1)(x - 3)$:
the coordinates of its $c$-intercept are
the coordinates of its $x$-intercepts are
the degree of $c$ is
as $x \to -\infty$, $f(x) \to$?
as $x \to \infty$, $f(x) \to$?
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Part 1: C - intercept (y - intercept)
Step 1: Recall the definition of C - intercept
The C - intercept (or y - intercept) of a function \( C(x) \) occurs where \( x = 0 \). So we substitute \( x=0 \) into the function \( C(x)=-(x - 5)(x + 1)(x - 3) \).
Step 2: Substitute \( x = 0 \) into the function
\[
\]
So the coordinates of the C - intercept are \( (0,-15) \).
Part 2: x - intercepts
Step 1: Recall the definition of x - intercepts
The x - intercepts of a function \( C(x) \) occur where \( C(x)=0 \). So we set \( -(x - 5)(x + 1)(x - 3)=0 \).
Step 2: Solve for x
Since the product of factors is zero, then at least one of the factors must be zero.
- For \( x - 5=0 \), we get \( x = 5 \), so the point is \( (5,0) \).
- For \( x + 1=0 \), we get \( x=-1 \), so the point is \( (-1,0) \).
- For \( x - 3=0 \), we get \( x = 3 \), so the point is \( (3,0) \).
Part 3: Degree of \( C(x) \)
Step 1: Recall the degree of a polynomial
The degree of a polynomial is the highest power of \( x \) when the polynomial is expanded. The function \( C(x)=-(x - 5)(x + 1)(x - 3) \) is a product of three linear factors (degree 1 each). When we multiply three linear factors, the degree of the resulting polynomial is the sum of the degrees of the factors. Since each factor has degree 1, the degree of the product is \( 1 + 1+1=3 \).
Part 4: End - behavior of the function
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s:
- The coordinates of its \( C \) - intercept are \( \boldsymbol{(0,-15)} \).
- The coordinates of its \( x \) - intercepts are \( \boldsymbol{(5,0)}, \boldsymbol{(-1,0)}, \boldsymbol{(3,0)} \).
- The Degree of \( C \) is \( \boldsymbol{3} \).
- As \( x
ightarrow-\infty \), \( f(x)
ightarrow\boldsymbol{\infty} \).
- As \( x
ightarrow\infty \), \( f(x)
ightarrow\boldsymbol{-\infty} \).