Question

A quadratic function has a line of symmetry at x = –3.5 and a zero at –9. What is the distance from the given zero to the line of symmetry?

Answer 1

Answer:

5.5. next one is 2

Step-by-step explanation:

just did it

Answer 2

The distance from the provided zero to the line of symmetry is 5.5 units after using the distance formula.

What is a quadratic equation?

Any equation of the form $\rm ax^2+bx+c=0$ where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

$\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}$

It is given that:

A quadratic function has a line of symmetry at x = –3.5 and a zero at –9.

As we know the standard form of the quadratic equation is:

$\rm ax^2+bx+c=0$

The points at the line of symmetry: (-3.5, 0)

The zero is at -9, so the coordinate point is: (-9, 0)

The distance between points (-3.5, 0) and (-9, 0)

Using the distance formula:

The distance formula can be given as:

$\rm d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\rm d=\sqrt{(-3.5+9)^2+(0-0)^2}$

d = 5.5 units

Thus, the distance from the provided zero to the line of symmetry is 5.5 units after using the distance formula.

Learn more about quadratic equations here:

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