Question

What is the vertex of the quadratic y = 6 ( x − 2 ) 2 + 10?

Quadratics
ANSWERED
asked 2021-02-22

What is the vertex of the quadratic \(y = 6 ( x − 2 )^2 + 10\)?

Answers (1)

2021-02-23
Given quadratic equation ,
\(\displaystyle{y}={6}{\left({x}-{2}\right)}^{{2}}+{10}\)
On simplifying,
\(\displaystyle{y}={6}{\left({x}^{{2}}+{4}-{4}{x}\right)}+{10}\)
\(\displaystyle{y}={6}{x}^{{2}}-{24}{x}+{24}+{10}\) (1)
\(\displaystyle{y}={6}{x}^{{2}}-{24}{x}+{34}\)
Now, this equation compare with general quadratic equation,
\(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}\)
a = 6 , b = -24 , c = 34
\(\displaystyle{h}=-\frac{{b}}{{2}}{a}\)
\(\displaystyle=-\frac{{-{24}}}{{{2}\cdot{6}}}\)
= 2
Now putting x= 2 in equation (i), then we get
\(\displaystyle{y}={6}\cdot{2}^{{2}}-{24}\cdot{2}+{34}\)
y = 24 - 48 + 34
y = 10
(2, 10) is the vertex of given quadratic equation.
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