Question

find the roots and the vertex of the quadratic on a calculator. round all values to 3 decimal places (if necessary).
y = -x² + 12x - 11
answer attempt 1 out of 2
roots:
and
vertex: (
,
)
submit answer

Explanation:

Step1: Find the roots using quadratic formula

For a quadratic equation \(ax^2 + bx + c = 0\), the roots are given by \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For \(y=-x^2 + 12x - 11\), set \(y = 0\), so \(-x^2+12x - 11 = 0\) or \(x^2-12x + 11 = 0\) (multiplying both sides by - 1). Here, \(a = 1\), \(b=-12\), \(c = 11\).

$$x=\frac{12\pm\sqrt{(-12)^2-4\times1\times11}}{2\times1}=\frac{12\pm\sqrt{144 - 44}}{2}=\frac{12\pm\sqrt{100}}{2}=\frac{12\pm10}{2}$$

Step2: Calculate the two roots

First root: \(\frac{12 + 10}{2}=\frac{22}{2}=11\)
Second root: \(\frac{12-10}{2}=\frac{2}{2}=1\)

Step3: Find the x - coordinate of the vertex

For a quadratic function \(y = ax^2+bx + c\), the x - coordinate of the vertex is \(x=-\frac{b}{2a}\). For \(y=-x^2 + 12x - 11\), \(a=-1\), \(b = 12\).

$$x=-\frac{12}{2\times(-1)}=\frac{-12}{-2}=6$$

Step4: Find the y - coordinate of the vertex

Substitute \(x = 6\) into the function \(y=-x^2+12x - 11\)

$$y=-(6)^2+12\times6-11=-36 + 72-11=25$$

Answer:

Roots: \(1\) and \(11\)
Vertex: \((6, 25)\)