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To sketch the graph of the quadratic function \( y = 2x^2 + 5x - 3 \), we need to determine the points where the graph crosses the coordinate axes (x-axis and y-axis).

### 1. Find the **y-intercept**:
The y-intercept occurs where \( x = 0 \).

\[
y = 2(0)^2 + 5(0) - 3 = -3
\]

Thus, the graph crosses the y-axis at \( (0, -3) \).

### 2. Find the **x-intercepts**:
The x-intercepts occur where \( y = 0 \). So, solve the quadratic equation:

\[
2x^2 + 5x - 3 = 0
\]

We can solve this using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 2 \), \( b = 5 \), and \( c = -3 \). Plugging these values into the quadratic formula:

\[
x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)}
\]
\[
x = \frac{-5 \pm \sqrt{25 + 24}}{4}
\]
\[
x = \frac{-5 \pm \sqrt{49}}{4}
\]
\[
x = \frac{-5 \pm 7}{4}
\]

So, the solutions are:

\[
x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}
\]
\[
x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3
\]

Thus, the graph crosses the x-axis at \( x = \frac{1}{2} \) and \( x = -3 \).

### 3. Vertex of the Parabola:
The vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:

\[
x_{\text{vertex}} = \frac{-b}{2a}
\]

Substituting the values \( a = 2 \) and \( b = 5 \):

\[
x_{\text{vertex}} = \frac{-5}{2(2)} = \frac{-5}{4}
\]

To find the y-coordinate of the vertex, substitute \( x = \frac{-5}{4} \) back into the original equation:

\[
y_{\text{vertex}} = 2\left( \frac{-5}{4} \right)^2 + 5\left( \frac{-5}{4} \right) - 3
\]
\[
y_{\text{vertex}} = 2\left( \frac{25}{16} \right) + 5\left( \frac{-5}{4} \right) - 3
\]
\[
y_{\text{vertex}} = \frac{50}{16} + \frac{-25}{4} - 3
\]
\[
y_{\text{vertex}} = \frac{50}{16} - \frac{100}{16} - \frac{48}{16} = \frac{50 - 100 - 48}{16} = \frac{-98}{16} = -6.125
\]

Thus, the vertex is at \( \left( \frac{-5}{4}, -6.125 \right) \).

### 4. Sketch:
- The graph is a parabola opening upwards because the coefficient of \( x^2 \) is positive.
- The graph crosses the y-axis at \( (0, -3) \).
- The graph crosses the x-axis at \( (-3, 0) \) and \( \left( \frac{1}{2}, 0 \right) \).
- The vertex of the parabola is at \( \left( \frac{-5}{4}, -6.125 \right) \).

Would you like to see a visual sketch or further details on any part of the solution?

Here are 5 related questions:
1. How do you determine the axis of symmetry for a parabola?
2. What does the discriminant tell us about the roots of a quadratic equation?
3. How can you factor a quadratic equation like \( 2x^2 + 5x - 3 \)?
4. What happens if the quadratic function has no real x-intercepts?
5. How do you calculate the focus and directrix of a parabola?

**Tip**: Always check the sign of the coefficient of \( x^2 \) in a quadratic equation to determine if the parabola opens upwards or downwards.

Sketch the graph of y = 2x^2 + 5x - 3, labelling all points where the graph crosses the coordinate axes.

Learn how to sketch the graph of the quadratic equation y = 2x^2 + 5x - 3, and identify the x- and y-intercepts, vertex, and the general shape of the parabola. Suitable for students in Grades 9-12.

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