https://file.aiquickdraw.com/mathbot/file/1727315319300x205vvfb.jpg

Let's solve this step by step, based on the given quadratic function \( y = 2x^2 + 6x - 3 \).

### a. Convert to vertex form \( y = a(x - h)^2 + k \)
The vertex form of a quadratic function is \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. To convert \( y = 2x^2 + 6x - 3 \) into vertex form, we complete the square.

1. Start with the original equation: 
   \[
   y = 2x^2 + 6x - 3
   \]

2. Factor out the coefficient of \( x^2 \) (which is 2) from the first two terms:
   \[
   y = 2(x^2 + 3x) - 3
   \]

3. Complete the square inside the parentheses. To do that, take half of the coefficient of \( x \) (which is 3), square it, and add/subtract it inside the parentheses. Half of 3 is \( \frac{3}{2} \), and squaring it gives \( \frac{9}{4} \). Add and subtract \( \frac{9}{4} \):
   \[
   y = 2\left(x^2 + 3x + \frac{9}{4} - \frac{9}{4}\right) - 3
   \]

4. Simplify the expression inside the parentheses:
   \[
   y = 2\left(\left(x + \frac{3}{2}\right)^2 - \frac{9}{4}\right) - 3
   \]

5. Distribute the 2:
   \[
   y = 2\left(x + \frac{3}{2}\right)^2 - 2 \times \frac{9}{4} - 3
   \]
   \[
   y = 2\left(x + \frac{3}{2}\right)^2 - \frac{9}{2} - 3
   \]

6. Combine the constants:
   \[
   y = 2\left(x + \frac{3}{2}\right)^2 - \frac{15}{2}
   \]

So, the vertex form is:
\[
y = 2\left(x + \frac{3}{2}\right)^2 - \frac{15}{2}
\]

### b. State the coordinates of the vertex
From the vertex form \( y = 2(x + \frac{3}{2})^2 - \frac{15}{2} \), the vertex is \( (h, k) \). Here, \( h = -\frac{3}{2} \) and \( k = -\frac{15}{2} \).

So, the vertex is:
\[
\left(-\frac{3}{2}, -\frac{15}{2}\right)
\]

### c. Find the y-intercept
The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the original equation:
\[
y = 2(0)^2 + 6(0) - 3 = -3
\]
Thus, the y-intercept is \( (0, -3) \).

### d. Sketch the graph of the function
The graph of the quadratic function will be a parabola opening upwards because the coefficient of \( x^2 \) (which is 2) is positive. The key points to note are:
- The vertex at \( \left(-\frac{3}{2}, -\frac{15}{2}\right) \)
- The y-intercept at \( (0, -3) \)
- Since \( a = 2 \), the parabola is narrower than the standard \( y = x^2 \).

Would you like me to sketch the graph for you?

---

Here are some related questions for deeper understanding:
1. How would the graph change if \( a \) were negative in the vertex form?
2. Can you explain how completing the square helps in finding the vertex form?
3. What would the x-intercepts be for this quadratic equation?
4. How does the value of \( a \) affect the "width" of the parabola?
5. How would you graph the function by plotting additional points around the vertex?

**Tip:** Completing the square is a very useful method for converting a quadratic equation into vertex form, as it directly reveals the vertex, making it easier to analyze and graph the function.

Consider the quadratic function y = 2x^2 + 6x - 3.
a. Convert it to the form y = a(x - h)^2 + k.
b. State the coordinates of the vertex.
c. Find the y-intercept.
d. Sketch the graph of the function.

Learn how to convert the quadratic function y = 2x^2 + 6x - 3 into vertex form using the method of completing the square. Find the vertex coordinates, y-intercept, and understand how to sketch the graph of the function. Suitable for students in Grades 9-11.

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