Answer:
The parabola that:
*opens up;
*has the vertex (2, 1);
*has no x-intercepts and;
*has the y-intercept 5.
Explanation:
The graph of function y = (x-2)²+1 looks like the picture attached.
How to get the information to determine what the graph looks like:
Recognize that the function is in vertex form y = a(x-h)² + k for parabolas.
"a" tells you the vertical stretch/compression of the parabola. Since it is not written in this equation, a=1. Since 1 is a positive number, the graph opens up.
The equation gives you the vertex of the parabola, which is the middle where the two curves meet. The vertex is a point in the form (x, y). The x-coordinate is "-h". Since h = -2, x = 2. The y-coordinate is "k". Since k = 1, y = 1. The vertex is (2, 1).
Since the graph opens up and the vertex is higher than the x-axis, there are no x-intercepts.
To find the y-intercept, expand the function.
y = (x-2)²+1 Write out both of the (x-2) in (x-2)²
y = (x-2)(x-2) + 1 Use FOIL or special products to expand
y = (x² - 4x + 4) + 1 Ignore the brackets, combine like terms (4 and 1)
y = x² - 4x + 5
The function, when expanded, is in standard form y = ax² + bx + c. The "c" variable in standard form tells you the y-intercept. Since c=5, the y-intercept is 5.