Question

given the function $f(x) = \

$$\begin{cases} x^2 + a & \\text{for } x \\leq 1 \\\\ -2ax + 7 & \\text{for } x > 1 \\end{cases}$$

$, what must the value of $a$ be so that $y = f(x)$ is continuous at $x = 1$? (you can draw it without lifting your pencil.)

Explanation:

Step1: Set left limit equal to right limit

For continuity at $x=1$, the value of the function from the left ($x\leq1$) must equal the value from the right ($x>1$) at $x=1$.
$$1^2 + a = -2a(1) + 7$$

Step2: Simplify the equation

Simplify both sides of the equality.
$$1 + a = -2a + 7$$

Step3: Solve for $a$

Isolate the variable $a$ by collecting like terms.
$$a + 2a = 7 - 1$$
$$3a = 6$$
$$a = \frac{6}{3} = 2$$

Answer:

$a=2$