Final answer:
After analyzing the given quadratic equations, the equation that does not have a vertex at (1, -2) is the first one, with a typo likely present. Once corrected, it shows a different vertex, not at (1, -2).
Step-by-step explanation:
To determine which of the given quadratic equations does not have a vertex at (1, -2), we need to analyze the equations and find the vertex form. A quadratic equation in vertex form is y = a(x - h)² + k, where (h, k) is the vertex of the parabola.
- For the equation y=x²-2x-1, rewriting it in vertex form gives us y = (x - 1)² - 2, with a vertex at (1, -2).
- In the case of y=1/4x²-1/2x-7/4, converting to vertex form yields y = 1/4(x - 1)² - 2, having the vertex at (1, -2) as well.
- Considering y= 3x²-6x+1, it simplifies to y = 3(x - 1)² - 2, resulting in a vertex at (1, -2).
- However, the first option y=1/4x²+3/2x=17 appears to have a typo or an extraneous component, rendering it unclear. After discarding the '=17' part, which is likely a typo, the remaining quadratic y = 1/4x² + 3/2x can be completed to y = 1/4(x + 3)² - 9/4, leading to a vertex at (-3, -9/4), distinctly not (1, -2).
Consequently, the equation that does not have the vertex at (1, -2) is the first one, provided we ignore the likely typo.