Answer :
Explanation:
To determine the zero of the linear polynomial given that the points \((- \frac{3}{2}, 0)\), \((-2, -1)\), and \((0, 3)\) lie on its graph, we need to understand that the zero of the polynomial is the x-coordinate where the value of the polynomial is zero.
Given the point \((- \frac{3}{2}, 0)\):
- This point means that when \( x = - \frac{3}{2} \), the value of the polynomial is 0.
- Therefore, the zero of the polynomial is \( x = - \frac{3}{2} \).
To verify, we can check if the given points lie on a linear polynomial \( y = mx + c \) and if they satisfy the condition for this polynomial having a zero at \( x = - \frac{3}{2} \).
1. **Point \((- \frac{3}{2}, 0)\)**:
- The polynomial is zero at \( x = - \frac{3}{2} \), so it must be true that:
\[
m \left( - \frac{3}{2} \right) + c = 0
\]
2. **Point \((-2, -1)\)**:
- The polynomial must satisfy:
\[
-1 = m \left( -2 \right) + c
\]
3. **Point \((0, 3)\)**:
- The polynomial must satisfy:
\[
3 = m(0) + c \implies c = 3
\]
Using \( c = 3 \) in the equation for point \((-2, -1)\):
\[
-1 = m(-2) + 3
\]
\[
-1 = -2m + 3
\]
\[
-1 - 3 = -2m
\]
\[
-4 = -2m
\]
\[
m = 2
\]
Thus, the linear polynomial is:
\[
y = 2x + 3
\]
We can verify if \((- \frac{3}{2}, 0)\) satisfies this polynomial:
\[
0 = 2 \left( - \frac{3}{2} \right) + 3
\]
\[
0 = -3 + 3
\]
\[
0 = 0
\]
So, the point \((- \frac{3}{2}, 0)\) lies on the polynomial, confirming that the zero of the polynomial is indeed:
\[
x = - \frac{3}{2}
\]
Answer:
Graph of a linear polynomial is a straight line. If points (- 3/2, 0), (- 2, - 1), (0, 3) lie on this graph of polynomial, write the zero of the polynomial.