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Find the vertex, the \(x\) -intercepts (if any), and sketch the parabola. \(f(x)=3 x^{2}-4 x+2\)

Short Answer

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The vertex of the parabola is \(\left(\dfrac{2}{3}, \dfrac{10}{3}\right)\). There are no \(x\)-intercepts, as the discriminant is negative. The parabola opens upwards since the leading coefficient is positive. The sketch of the parabola includes the vertex and points at \((1,1)\) and \(\left(\dfrac{1}{3},1\right)\) with no intersections on the \(x\)-axis.
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Step 1: Find the Vertex

The vertex of a quadratic function \(f(x) = ax^2 + bx + c\) is given by the formula \(x = \dfrac{-b}{2a}\). First, let's find the \(x\)-coordinate of the vertex by plugging the values of \(a\) and \(b\): \(x = \dfrac{-(-4)}{2(3)} = \dfrac{4}{6} = \dfrac{2}{3}\) Now, to find the \(y\)-coordinate of the vertex, plug the \(x\)-coordinate into the function: \(f\left(\dfrac{2}{3}\right) = 3\left(\dfrac{2}{3}\right)^2 - 4\left(\dfrac{2}{3}\right) + 2\) \(f\left(\dfrac{2}{3}\right) = 4 - \dfrac{8}{3} + 2 = \dfrac{10}{3}\) So, the vertex of the parabola is \(\left(\dfrac{2}{3}, \dfrac{10}{3}\right)\).

Step 2: Find the x-intercepts

To find the \(x\)-intercepts, we need to solve the equation \(f(x) = 0\): \(3x^2 - 4x + 2 = 0\) Since this quadratic equation doesn't factor easily, we'll use the quadratic formula: \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Plug in \(a=3\), \(b=-4\), and \(c=2\): \(x = \dfrac{4 \pm \sqrt{(-4)^2 - 4(3)(2)}}{2(3)}\) \(x = \dfrac{4 \pm \sqrt{16 - 24}}{6}\) \(x = \dfrac{4 \pm \sqrt{-8}}{6}\) Since there's a negative value under the square root, there are no real solutions for \(x\). Thus, there are no \(x\)-intercepts.

Step 3: Sketch the Parabola

Now that we have the vertex and know that there are no \(x\)-intercepts, we can sketch the parabola: 1. Plot the vertex at \(\left(\dfrac{2}{3}, \dfrac{10}{3}\right)\). 2. Since \(a > 0\), the parabola opens upwards. 3. Note that there are no \(x\)-intercepts, so the parabola does not touch the \(x\)-axis. 4. Use one more point to determine the shape of the parabola. For example, find the value of \(f(1)\): \(f(1) = 3(1)^2 - 4(1) + 2 = 1\) Now, plot this point \((1,1)\). 5. Reflect the point \((1,1)\) over the vertical line \(x=\dfrac{2}{3}\) to find a symmetric point on the other side of the parabola, which is \(\left(\dfrac{1}{3},1\right)\). 6. Connect the points with the curve representing the parabola. Now, you have sketched the parabola for \(f(x) = 3x^2 - 4x + 2\) with vertex \(\left(\dfrac{2}{3}, \dfrac{10}{3}\right)\) and no \(x\)-intercepts.

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