Question
Use a graphing utility as an aid in factoring to solve the equation.
3xsquaredplus14xminus49equals0
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Question content area right
Part 1
Choose the correct graph of the equation.
A.
A coordinate system has a horizontal axis from negative 10 to 10 in increments of 2 and a vertical axis from negative 70 to 20 in increments of 10. A parabola that opens upward has a vertex in the third quadrant.
[minus10,10] by [negative 70,20] with Xsclequals2 and Ysclequals10
B.
A coordinate system has a horizontal axis from negative 10 to 10 in increments of 2 and a vertical axis from negative 70 to 20 in increments of 10. A parabola that opens upward has a vertex in the fourth quadrant.
[minus10,10] by [negative 70,20] with Xsclequals2 and Ysclequals10
C.
A coordinate system has a horizontal axis from negative 10 to 10 in increments of 2 and a vertical axis from negative 20 to negative negative 70 in increments of 10. A parabola that opens downward has a vertex in the second quadrant.
[minus10,10] by [-20,70] with Xsclequals2 and Ysclequals10
D.
A coordinate system has a horizontal axis from negative 10 to 10 in increments of 2 and a vertical axis from negative 20 to negative negative 70 in increments of 10. A parabola that opens downward has a vertex in the first quadrant.
[minus10,10] by [-20,70] with Xsclequals2 and Ysclequals10
3xsquaredplus14xminus49equals0
.
.
.
Question content area right
Part 1
Choose the correct graph of the equation.
A.
A coordinate system has a horizontal axis from negative 10 to 10 in increments of 2 and a vertical axis from negative 70 to 20 in increments of 10. A parabola that opens upward has a vertex in the third quadrant.
[minus10,10] by [negative 70,20] with Xsclequals2 and Ysclequals10
B.
A coordinate system has a horizontal axis from negative 10 to 10 in increments of 2 and a vertical axis from negative 70 to 20 in increments of 10. A parabola that opens upward has a vertex in the fourth quadrant.
[minus10,10] by [negative 70,20] with Xsclequals2 and Ysclequals10
C.
A coordinate system has a horizontal axis from negative 10 to 10 in increments of 2 and a vertical axis from negative 20 to negative negative 70 in increments of 10. A parabola that opens downward has a vertex in the second quadrant.
[minus10,10] by [-20,70] with Xsclequals2 and Ysclequals10
D.
A coordinate system has a horizontal axis from negative 10 to 10 in increments of 2 and a vertical axis from negative 20 to negative negative 70 in increments of 10. A parabola that opens downward has a vertex in the first quadrant.
[minus10,10] by [-20,70] with Xsclequals2 and Ysclequals10
Answers
Answered by
GPT-4o mini
To solve the equation \(3x^2 + 14x - 49 = 0\) by factoring, we can first determine its roots through graphing. However, let's rearrange and analyze the quadratic to decide how it behaves.
1. **Identify the vertex and direction**:
The coefficient of \(x^2\) is positive (3), indicating that the parabola opens upwards.
2. **Finding the vertex**:
The formula for the x-coordinate of the vertex of a quadratic equation \(ax^2 + bx + c\) is given by:
\[
x = -\frac{b}{2a}
\]
For our equation:
- \(a = 3\)
- \(b = 14\)
So,
\[
x = -\frac{14}{2 \cdot 3} = -\frac{14}{6} = -\frac{7}{3} \approx -2.33
\]
To find the y-coordinate of the vertex, substitute \(x = -\frac{7}{3}\) into the equation:
\[
y = 3\left(-\frac{7}{3}\right)^2 + 14\left(-\frac{7}{3}\right) - 49
\]
Simplifying this:
\[
= 3\left(\frac{49}{9}\right) - \frac{98}{3} - 49
\]
\[
= \frac{147}{9} - \frac{294}{9} - \frac{441}{9}
\]
\[
= \frac{147 - 294 - 441}{9} = \frac{-588}{9} \approx -65.33
\]
So, the vertex is approximately at \((-2.33, -65.33)\), which is located in the third quadrant.
3. **Examine the choices**:
Given that the parabola opens upwards and the vertex is in the third quadrant, we can analyze the options:
- **Option A** indicates a vertex in the third quadrant and a parabola opening upwards.
- **Option B** has a vertex in the fourth quadrant (incorrect).
- **Option C** and **D** show downward-opening parabolas (incorrect).
Thus, the correct graph of the equation is **Option A**.
1. **Identify the vertex and direction**:
The coefficient of \(x^2\) is positive (3), indicating that the parabola opens upwards.
2. **Finding the vertex**:
The formula for the x-coordinate of the vertex of a quadratic equation \(ax^2 + bx + c\) is given by:
\[
x = -\frac{b}{2a}
\]
For our equation:
- \(a = 3\)
- \(b = 14\)
So,
\[
x = -\frac{14}{2 \cdot 3} = -\frac{14}{6} = -\frac{7}{3} \approx -2.33
\]
To find the y-coordinate of the vertex, substitute \(x = -\frac{7}{3}\) into the equation:
\[
y = 3\left(-\frac{7}{3}\right)^2 + 14\left(-\frac{7}{3}\right) - 49
\]
Simplifying this:
\[
= 3\left(\frac{49}{9}\right) - \frac{98}{3} - 49
\]
\[
= \frac{147}{9} - \frac{294}{9} - \frac{441}{9}
\]
\[
= \frac{147 - 294 - 441}{9} = \frac{-588}{9} \approx -65.33
\]
So, the vertex is approximately at \((-2.33, -65.33)\), which is located in the third quadrant.
3. **Examine the choices**:
Given that the parabola opens upwards and the vertex is in the third quadrant, we can analyze the options:
- **Option A** indicates a vertex in the third quadrant and a parabola opening upwards.
- **Option B** has a vertex in the fourth quadrant (incorrect).
- **Option C** and **D** show downward-opening parabolas (incorrect).
Thus, the correct graph of the equation is **Option A**.