Answered

Find the axis of symmetry for the following quadratic equation:

[tex]\[ y = -x^2 + 3x + 10 \][/tex]



Answer :

To find the axis of symmetry for the quadratic equation [tex]\( y = -x^2 + 3x + 10 \)[/tex], follow these steps:

1. Identify the coefficients:
The general form of a quadratic equation is [tex]\( y = ax^2 + bx + c \)[/tex]. For the given equation [tex]\( y = -x^2 + 3x + 10 \)[/tex], we can identify the coefficients as follows:
- [tex]\( a = -1 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( c = 10 \)[/tex]

2. Use the formula for the axis of symmetry:
The formula to find the axis of symmetry for a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] is [tex]\( x = -\frac{b}{2a} \)[/tex].

3. Substitute the coefficients into the formula:
Substitute [tex]\( a = -1 \)[/tex] and [tex]\( b = 3 \)[/tex] into the formula:
[tex]\[ x = -\frac{3}{2(-1)} \][/tex]

4. Simplify the expression:
Simplify the fraction:
[tex]\[ x = -\frac{3}{-2} \][/tex]

5. Calculate the result:
Simplifying further gives:
[tex]\[ x = \frac{3}{2} \][/tex]

So, the axis of symmetry for the quadratic equation [tex]\( y = -x^2 + 3x + 10 \)[/tex] is [tex]\( x = 1.5 \)[/tex]. This vertical line [tex]\( x = 1.5 \)[/tex] represents the axis of symmetry for the parabola described by the given quadratic equation.