The equation $y = 2x^2 - 8x + 3$ represents a quadratic function. The graph of a quadratic function is always a parabola. In this case, since the coefficient of $x^2$ is positive ($2$), the parabola will open upward.

To describe the graph in detail, we can follow these steps:

1. Vertex of the Parabola:

The vertex form of a quadratic equation is given by $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola. For the equation $y = 2x^2 - 8x + 3$, we can find the vertex using the formula for the x-coordinate of the vertex: $x = \frac{-b}{2a}$ For the given equation, $a = 2$ and $b = -8$. Substituting these values: $x = \frac{-(-8)}{2(2)} = \frac{8}{4} = 2$ Now substitute $x = 2$ back into the original equation to find the corresponding y-coordinate: $y = 2(2)^2 - 8(2) + 3 = 2(4) - 16 + 3 = 8 - 16 + 3 = -5$ Thus, the vertex is $(2, -5)$.

2. Axis of Symmetry:

The axis of symmetry is the vertical line that passes through the vertex. Since the x-coordinate of the vertex is 2, the axis of symmetry is the line $x = 2$.

3. Direction of the Parabola:

As mentioned earlier, since the coefficient of $x^2$ is positive ($a = 2$), the parabola opens upward.

4. Y-Intercept:

To find the y-intercept, set $x = 0$ and solve for $y$: $y = 2(0)^2 - 8(0) + 3 = 3$ Thus, the y-intercept is $(0, 3)$.

5. X-Intercepts:

To find the x-intercepts, set $y = 0$ and solve for $x$: $0 = 2x^2 - 8x + 3$ This is a quadratic equation, and we can solve it using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $2x^2 - 8x + 3 = 0$, $a = 2$, $b = -8$, and $c = 3$. Substituting these values: $x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(2)(3)}}{2(2)} = \frac{8 \pm \sqrt{64 - 24}}{4} = \frac{8 \pm \sqrt{40}}{4} = \frac{8 \pm 2\sqrt{10}}{4}$ Simplifying: $x = \frac{8}{4} \pm \frac{2\sqrt{10}}{4} = 2 \pm \frac{\sqrt{10}}{2}$ Thus, the x-intercepts are: $x = 2 + \frac{\sqrt{10}}{2} \quad \text{and} \quad x = 2 - \frac{\sqrt{10}}{2}$

Summary:

• The vertex of the parabola is $(2, -5)$.
• The axis of symmetry is $x = 2$.
• The parabola opens upward.
• The y-intercept is $(0, 3)$.
• The x-intercepts are $x = 2 + \frac{\sqrt{10}}{2}$ and $x = 2 - \frac{\sqrt{10}}{2}$.

Would you like to explore the graph visually, or do you have any specific questions about these steps?

Here are some related questions:

1. How can we transform this quadratic equation to its vertex form?
2. How do we identify whether a parabola opens upward or downward based on the equation?
3. How can the discriminant help us determine the number of real roots for a quadratic equation?
4. What are the steps to graph any quadratic function?
5. How does the value of $a$ in $ax^2 + bx + c$ affect the shape of the parabola?

Tip: The vertex is always the point that represents the minimum or maximum value of the quadratic function, depending on the direction the parabola opens.