Select The Correct Answer.Which Point Is A Point Where The Graph Of Y=(x-5)\left(x^2-7x+12\right ] Crosses The X X X -axis?A. ( − 5 , 0 (-5,0 ( − 5 , 0 ] B. ( − 3 , 0 (-3,0 ( − 3 , 0 ] C. ( 4 , 0 (4,0 ( 4 , 0 ] D. ( 12 , 0 (12,0 ( 12 , 0 ]

Understanding the Problem

To find the points where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis, we need to find the values of $x$ for which the function $y$ equals zero. This is because the $x$-axis is defined by the equation $y=0$. Therefore, we are looking for the values of $x$ that make the function $y$ equal to zero.

Breaking Down the Function

The function $y=(x-5)\left(x^2-7x+12\right)$ can be broken down into two parts: $(x-5)$ and $\left(x^2-7x+12\right)$. To find the values of $x$ that make the function equal to zero, we need to find the values of $x$ that make each of these parts equal to zero.

Finding the Values of $x$

To find the values of $x$ that make the part $(x-5)$ equal to zero, we can set it equal to zero and solve for $x$. This gives us the equation $x-5=0$, which has the solution $x=5$.

To find the values of $x$ that make the part $\left(x^2-7x+12\right)$ equal to zero, we can set it equal to zero and solve for $x$. This gives us the quadratic equation $x^2-7x+12=0$. We can solve this equation using the quadratic formula, which is given by $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. In this case, $a=1$, $b=-7$, and $c=12$. Plugging these values into the quadratic formula, we get $x=\frac{-(-7)\pm\sqrt{(-7)^2-4(1)(12)}}{2(1)}=\frac{7\pm\sqrt{49-48}}{2}=\frac{7\pm\sqrt{1}}{2}=\frac{7\pm1}{2}$. This gives us two possible values for $x$: $x=\frac{7+1}{2}=4$ and $x=\frac{7-1}{2}=3$.

Finding the Points Where the Graph Crosses the $x$-Axis

Now that we have found the values of $x$ that make each part of the function equal to zero, we can find the points where the graph of the function crosses the $x$-axis. To do this, we need to find the values of $x$ that make the entire function equal to zero. This means that we need to find the values of $x$ that make both parts of the function equal to zero.

The only value of $x$ that makes both parts of the function equal to zero is $x=5$. This is because $x=5$ makes the part $(x-5)$ equal to zero, and it also makes the part $\left(x^2-7x+12\right)$ equal to zero.

Conclusion

In conclusion, the point where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis is $(5,0)$. This is because $x=5$ makes the entire function equal to zero, which means that the graph of the function crosses the $x$-axis at this point.

Answer

The correct answer is A. $(5,0)$.

However, since this answer is not among the options provided, we need to re-evaluate the problem. Let's go back to the quadratic equation $x^2-7x+12=0$. We can factor this equation as $(x-3)(x-4)=0$. This gives us two possible values for $x$: $x=3$ and $x=4$.

Re-Evaluating the Problem

Now that we have factored the quadratic equation, we can re-evaluate the problem. We are looking for the values of $x$ that make the function $y=(x-5)\left(x^2-7x+12\right)$ equal to zero. This means that we need to find the values of $x$ that make both parts of the function equal to zero.

The values of $x$ that make the part $(x-5)$ equal to zero are $x=5$. The values of $x$ that make the part $\left(x^2-7x+12\right)$ equal to zero are $x=3$ and $x=4$.

Finding the Points Where the Graph Crosses the $x$-Axis

Now that we have found the values of $x$ that make each part of the function equal to zero, we can find the points where the graph of the function crosses the $x$-axis. To do this, we need to find the values of $x$ that make the entire function equal to zero. This means that we need to find the values of $x$ that make both parts of the function equal to zero.

The values of $x$ that make both parts of the function equal to zero are $x=3$ and $x=4$. Therefore, the points where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis are $(3,0)$ and $(4,0)$.

Conclusion

In conclusion, the points where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis are $(3,0)$ and $(4,0)$.

Answer

The correct answers are B. $(3,0)$ and C. $(4,0)$.

Q: What is the main goal of solving for the points where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis?

A: The main goal of solving for the points where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis is to find the values of $x$ for which the function $y$ equals zero. This is because the $x$-axis is defined by the equation $y=0$.

Q: How do we break down the function $y=(x-5)\left(x^2-7x+12\right)$ to solve for the points where it crosses the $x$-axis?

A: We can break down the function $y=(x-5)\left(x^2-7x+12\right)$ into two parts: $(x-5)$ and $\left(x^2-7x+12\right)$. To find the values of $x$ that make the function equal to zero, we need to find the values of $x$ that make each of these parts equal to zero.

Q: How do we find the values of $x$ that make the part $(x-5)$ equal to zero?

A: To find the values of $x$ that make the part $(x-5)$ equal to zero, we can set it equal to zero and solve for $x$. This gives us the equation $x-5=0$, which has the solution $x=5$.

Q: How do we find the values of $x$ that make the part $\left(x^2-7x+12\right)$ equal to zero?

A: To find the values of $x$ that make the part $\left(x^2-7x+12\right)$ equal to zero, we can set it equal to zero and solve for $x$. This gives us the quadratic equation $x^2-7x+12=0$. We can solve this equation using the quadratic formula, which is given by $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. In this case, $a=1$, $b=-7$, and $c=12$. Plugging these values into the quadratic formula, we get $x=\frac{-(-7)\pm\sqrt{(-7)^2-4(1)(12)}}{2(1)}=\frac{7\pm\sqrt{49-48}}{2}=\frac{7\pm\sqrt{1}}{2}=\frac{7\pm1}{2}$. This gives us two possible values for $x$: $x=\frac{7+1}{2}=4$ and $x=\frac{7-1}{2}=3$.

Q: How do we find the points where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis?

A: To find the points where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis, we need to find the values of $x$ that make the entire function equal to zero. This means that we need to find the values of $x$ that make both parts of the function equal to zero.

Q: What are the points where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis?

A: The points where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis are $(3,0)$ and $(4,0)$.

Q: Why are these points the correct answers?

A: These points are the correct answers because they are the values of $x$ that make the entire function equal to zero. This means that they are the points where the graph of the function crosses the $x$-axis.

Q: What is the final answer to the problem?

A: The final answer to the problem is that the points where the graph of $y=(x-5)\left(x^2-7x+12\right)$ crosses the $x$-axis are $(3,0)$ and $(4,0)$.