Questions: Find the (x)-and (y)-intercepts of the graph of (f(x)=(9x+5)(x-7)(3x-2)). State the degree and maximum possible number of turning points that the graph could have.

Find the (x)-and (y)-intercepts of the graph of (f(x)=(9x+5)(x-7)(3x-2)). State the degree and maximum possible number of turning points that the graph could have.

Transcript text: Question 12 Mark this question Find the $x$-and $y$-intercepts of the graph of $f(x)=(9 x+5)(x-7)(3 x-2)$. State the degree and maximum possible number of turning points that the graph could have.

Solution

Solution Steps

To find the $x$-intercepts of the function $f(x) = (9x + 5)(x - 7)(3x - 2)$, we need to set $f(x) = 0$ and solve for $x$. The $y$-intercept is found by evaluating $f(0)$. The degree of the polynomial is the sum of the degrees of its factors, and the maximum number of turning points is one less than the degree.

Step 1: Find the $x$-intercepts

To find the $x$-intercepts of the function $f(x) = (9x + 5)(x - 7)(3x - 2)$, we set $f(x) = 0$ and solve for $x$: \[ (9x + 5)(x - 7)(3x - 2) = 0 \] Solving each factor for zero, we get: \[ 9x + 5 = 0 \implies x = -\frac{5}{9} \] \[ x - 7 = 0 \implies x = 7 \] \[ 3x - 2 = 0 \implies x = \frac{2}{3} \] Thus, the $x$-intercepts are: \[ x = -\frac{5}{9}, \quad x = 7, \quad x = \frac{2}{3} \]

Step 2: Find the $y$-intercept

To find the $y$-intercept, we evaluate $f(0)$: \[ f(0) = (9 \cdot 0 + 5)(0 - 7)(3 \cdot 0 - 2) = 5 \cdot (-7) \cdot (-2) = 70 \] Thus, the $y$-intercept is: \[ y = 70 \]

Step 3: Determine the degree of the polynomial

The degree of the polynomial $f(x) = (9x + 5)(x - 7)(3x - 2)$ is the sum of the degrees of its factors. Each factor is a linear polynomial (degree 1), so the degree of $f(x)$ is: \[ 1 + 1 + 1 = 3 \]

Step 4: Determine the maximum number of turning points

The maximum number of turning points of a polynomial is one less than its degree. Since the degree of $f(x)$ is 3, the maximum number of turning points is: \[ 3 - 1 = 2 \]