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