Question

find the roots and the vertex of the quadratic on a calculator. round all values to 3 decimal places (if necessary).
y = -x² + 10x + 24
answer attempt 1 out of 2
roots:
and
vertex: (
,
)

Explanation:

Step1: Find the roots by solving \(-x^2 + 10x + 24 = 0\)

Multiply both sides by -1: \(x^2 - 10x - 24 = 0\)
Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b=-10\), \(c=-24\)
Discriminant: \(b^2 - 4ac=(-10)^2 - 4(1)(-24)=100 + 96 = 196\)
\(x=\frac{10\pm\sqrt{196}}{2}=\frac{10\pm14}{2}\)

Step2: Calculate the two roots

First root: \(\frac{10 + 14}{2}=\frac{24}{2}=12\)
Second root: \(\frac{10 - 14}{2}=\frac{-4}{2}=-2\)

Step3: Find the vertex of the quadratic \(y = ax^2+bx + c\)

The x - coordinate of the vertex is \(x=-\frac{b}{2a}\), for \(y=-x^2 + 10x + 24\), \(a=-1\), \(b = 10\)
\(x=-\frac{10}{2\times(-1)}=\frac{-10}{-2}=5\)

Step4: Find the y - coordinate of the vertex

Substitute \(x = 5\) into \(y=-x^2+10x + 24\)
\(y=-(5)^2+10\times5 + 24=-25 + 50+24=49\)

Answer:

Roots: \(-2\) and \(12\)
Vertex: \((5, 49)\)