Question

can someone help me with this math hw ​

Answer 1

Answer:

Step-by-step explanation:

F(x) = x² - 2x + 1

      = (x - 1)²

By comparing this equation with the vertex form of the quadratic equation,

y = (x - h)² + k

Here, (h, k) is the vertex

Vertex of the parabola → (1, 0)

x-intercepts → (x - 1)² = 0

                       x = 1

y-intercepts → y = (0 - 1)²

                       y = 1

Now we can draw the graph of the given function,

From this graph,

As x → 0,

$\lim_{x \to 0^{-}} (x-1)^2=1$

$\lim_{x \to 0^{+}} (x-1)^2=1$

f(0) = (0 - 1)²

     = 1

Since, $\lim_{x \to 0^{-}} (x-1)^2=\lim_{x \to 0^{+}} (x-1)^2=1$

Therefore, given function is continuous at x = 0.