To write the given quadratic equation $y = -x^2 + 3x + 4$ in vertex form, we need to rewrite it in the form:

$y = a(x - h)^2 + k$

where $(h, k)$ is the vertex of the parabola.

Let's follow the steps to complete the square:

Step 1: Factor out the coefficient of $x^2$ from the first two terms.

The given equation is: $y = -x^2 + 3x + 4$ Factor out the negative sign from the first two terms: $y = -\left(x^2 - 3x\right) + 4$

Step 2: Complete the square inside the parentheses.

To complete the square, take half of the coefficient of $x$ (which is -3), square it, and add and subtract that value inside the parentheses. $\left(\frac{-3}{2}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$

Now add and subtract $\frac{9}{4}$ inside the parentheses: $y = -\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) + 4$

Step 3: Simplify the equation.

Now group the perfect square trinomial and simplify: $y = -\left(\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right) + 4$

Distribute the negative sign: $y = -\left(x - \frac{3}{2}\right)^2 + \frac{9}{4} + 4$

Convert 4 into a fraction with the same denominator as $\frac{9}{4}$: $y = -\left(x - \frac{3}{2}\right)^2 + \frac{9}{4} + \frac{16}{4}$

Now add the fractions: $y = -\left(x - \frac{3}{2}\right)^2 + \frac{25}{4}$

Final Vertex Form:

The vertex form of the equation is: $y = -\left(x - \frac{3}{2}\right)^2 + \frac{25}{4}$

Here, the vertex of the parabola is $\left(\frac{3}{2}, \frac{25}{4}\right)$.

Would you like further details on the steps or have any questions?

Related Questions:

1. How do you find the vertex of a quadratic equation from standard form?
2. What is the significance of the vertex in a parabola?
3. How does the sign of the leading coefficient affect the direction of the parabola?
4. How do you convert a quadratic equation in vertex form back to standard form?
5. Can all quadratic equations be converted to vertex form using completing the square?

Tip:

When completing the square, always remember to balance your equation by both adding and subtracting the same value.