To convert the equation $y = -4(x-1)(2x-5)$ into vertex form, we need to first expand and simplify it, and then complete the square. Here's the step-by-step process:

Step 1: Expand the equation

First, expand the product of the binomials $(x-1)(2x-5)$.

$(x-1)(2x-5) = x(2x-5) - 1(2x-5)$ $= 2x^2 - 5x - 2x + 5$ $= 2x^2 - 7x + 5$

Now substitute this back into the equation for $y$:

$y = -4(2x^2 - 7x + 5)$

Step 2: Distribute the $-4$

Next, distribute the $-4$ across the terms inside the parentheses:

$y = -8x^2 + 28x - 20$

Step 3: Complete the square

To express the equation in vertex form, $y = a(x-h)^2 + k$, we need to complete the square. First, factor out the $-8$ from the quadratic and linear terms:

$y = -8(x^2 - \frac{7}{4}x) - 20$

Now, to complete the square, we need to add and subtract a value inside the parentheses that makes $x^2 - \frac{7}{4}x$ a perfect square trinomial. The value we need to add is $\left(\frac{7}{8}\right)^2 = \frac{49}{64}$.

$y = -8\left(x^2 - \frac{7}{4}x + \frac{49}{64} - \frac{49}{64}\right) - 20$

$y = -8\left[\left(x - \frac{7}{8}\right)^2 - \frac{49}{64}\right] - 20$

Step 4: Simplify

Now, distribute the $-8$ to both terms inside the brackets:

$y = -8\left(x - \frac{7}{8}\right)^2 + \frac{392}{64} - 20$

Convert $\frac{392}{64}$ to a simpler fraction:

$\frac{392}{64} = 6.125$

Now, subtract 20 from $6.125$:

$y = -8\left(x - \frac{7}{8}\right)^2 - 13.875$

Thus, the equation in vertex form is:

$y = -8\left(x - \frac{7}{8}\right)^2 - 13.875$

Vertex

The vertex of the parabola is at $\left(\frac{7}{8}, -13.875\right)$.

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Related Questions:

1. How do you convert from standard form to vertex form?
2. What is the significance of completing the square in parabolas?
3. How can you find the axis of symmetry from vertex form?
4. What is the difference between vertex form and factored form?
5. How do you determine whether a parabola opens upwards or downwards?

Tip:

When completing the square, always remember to add and subtract the same value to maintain the equality of the equation.