Algebra 1, Please help! Vertex Form

Question

Given the function f(x)=(x+3)2−1, what is the location of the vertex? 

1. (3,1)
2. (-3,1)
3. (3,-1)
4. (-3,-1)

Given the function f(x)=(x+3)2−1, does the graph open up or down?

The graph representing y=x2 (in blue) is shifted 2 units to the left, 10 units down, and flipped so that it opens downward, as shown (in black). Write the equation, in vertex form, that defines the shifted curve. Explain the values of a, h, and k in your new equation.

y=a(x−h)²+k

Graph Image: https://assets.learnosity.com/organisations/154/888fa567-b096-4a11-8d28-3b5ed91b169c.PNG

Answer 1

Vertex form: f(x) = y = a(x - h)² + k where "a" is the coefficient of the 1st term of the quadratic in std. form, "h" is the x-coordinate of the vertex and "k" is the y-coordinate of the vertex. So for f(x) = (x + 3)² - 1, "h" is -3 and "k" is -1, so answer to the location of the vertex is (-3,-1).

It opens upwards, because the "a" is > 0. (= 1 in this case).

re: y = x², let's do it in steps. To shift 2 units left, need do the opposite to the x term (add 2), so function becomes y = (x + 2)². To shift 10 units down is just as it seems; subtract 10 from the function: y = (x + 2)² - 10. To flip it so it opens downwards, change the "a" term to its negative: y = -(x + 2)² - 10. New vertex is at (-2, -10) and opens downwards.

Try these on the Desmos Graphing Calculator website to see the effect of the changes in each step.