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To solve the equation [tex]-x-3=x^2-2x-15[/tex], you could graph the following system:

[tex]
\begin{array}{l}
y = -x - 3 \\
y = x^2 - 2x - 15
\end{array}
[/tex]

Use the graph of this system to identify all solutions of [tex]-x-3=x^2-2x-15[/tex].

A. -7
B. -3
C. 0
D. 4

Sagot :

GinnyAnswer
To solve the equation [tex]\(-x - 3 = x^2 - 2x - 15\)[/tex] by using the graph of the system:

[tex]\[ \begin{array}{l} y = -x - 3 \\ y = x^2 - 2x - 15 \end{array} \][/tex]

you need to find the points where the two graphs intersect. Here’s a step-by-step process to understand and find the solutions:

1. Graph the Linear Equation:
- The equation [tex]\(y = -x - 3\)[/tex] is a linear equation.
- The slope of this line is [tex]\(-1\)[/tex] and the y-intercept is [tex]\(-3\)[/tex]. This means the line crosses the y-axis at [tex]\(-3\)[/tex] and goes downwards with a slope of [tex]\(-1\)[/tex].

2. Graph the Quadratic Equation:
- The equation [tex]\(y = x^2 - 2x - 15\)[/tex] is a quadratic equation, which forms a parabola.
- The parabola opens upwards because the coefficient of [tex]\(x^2\)[/tex] is positive.
- To better graph this, you would typically find the vertex and the x-intercepts (roots) of the parabola. However, for the intersection points, we can focus on the points where it meets the linear graph.

3. Find Intersection Points:
- To find the exact points where these graphs intersect, we set the equations equal to each other: [tex]\(-x - 3 = x^2 - 2x - 15\)[/tex].
- This results in the equation: [tex]\(x^2 - x - 12 = 0\)[/tex].

4. Solve the Quadratic Equation:
- The quadratic equation [tex]\(x^2 - x - 12 = 0\)[/tex] can be factored as:
[tex]\[ (x - 4)(x + 3) = 0 \][/tex]
- This gives the solutions:
[tex]\[ x - 4 = 0 \quad \text{or} \quad x + 3 = 0 \][/tex]
[tex]\[ x = 4 \quad \text{or} \quad x = -3 \][/tex]

5. Confirm the Points:
- Substitute these [tex]\(x\)[/tex]-values back into either of the original equations to confirm the intersection points.
- For [tex]\(x = 4\)[/tex]:
[tex]\[ y = -4 - 3 = -7 \][/tex]
- For [tex]\(x = -3\)[/tex]:
[tex]\[ y = 3 - 3 = 0 \][/tex]

Hence, the intersection points of the graphs are [tex]\((-3, 0)\)[/tex] and [tex]\((4, -7)\)[/tex]. Thus, the solutions to the equation [tex]\(-x - 3 = x^2 - 2x - 15\)[/tex] are:

[tex]\[ x = -3 \quad \text{and} \quad x = 4. \][/tex]

Therefore, the solutions identified from the graph are:
- [tex]\(-3\)[/tex]
- [tex]\(4\)[/tex]
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