Question

A quadratic function has a line of symmetry at x = –3 and a zero at 4.

Answer 1

I have my hints here on how to solve the questions one and two:

Q1: Just find the distance from -3 to 4. That's 7.

Q2: It is 7 units to the left of -3. Subtract 7 from -3. That's -10

I hope my answer has come to your help. God bless and have a nice day ahead!

Answer 2

Answer:

Part 1) $7\ units$

Part 2) The other zero is $-10$

Step-by-step explanation:

we know that

If the line of symmetry is equal to $x=-3$

then

the quadratic function is a vertical parabola and the x-coordinate of the vertex is $-3$

Part 1) What is the distance from the given zero to the line of symmetry?

The distance is equal to the absolute value of the difference of $-3$ from $4$

$d=\left|4-\left(-3\right)\right|=7\ units$  

Part 2) What is the other zero of the quadratic function?

we know that

In a quadratic function the distance from the zero's to the line of symmetry is equal

so

Subtract $7$ from $-3$

$-3-7=-10$

The other zero is $-10$