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Problem 10

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

Expert verified
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.
See the step by step solution

## 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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