How do you find the vertex of $f (x) = 3 x^{2} - 25$ $f(x)=3x^2-25$ ?

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Answer 1 · 2016-10-07T19:55:47

$V e r t e x \Rightarrow (x, y) \Rightarrow (0, - 25)$ $Vertex => (x,y) => (0,-25)$

Standard form of a quadratic equation

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$x = - \frac{b}{2} a$ $x=-b/2a$

$y = f (- \frac{b}{2 a})$ $y=f(-b/(2a))$

$V e r t e x \Rightarrow (x, y) \Rightarrow (- \frac{b}{2 a}, f (- \frac{b}{2 a}))$ $Vertex => (x,y) => (-b/(2a),f(-b/(2a)))$

$f (x) = 3 x^{2} - 25 = 3 x^{2} + 0 x - 25$ $f(x)=3x^2-25=3x^2+0x-25$

$a = 3$ $a=3$
$b = 0$ $b=0$
$c = - 25$ $c=-25$

Substitute

$x = - \frac{b}{2 a} = - \frac{0}{2 (3)} = 0$ $x=-b/(2a)=-0/(2(3))=0$

$y = f (- \frac{b}{2 a}) = f (0) = 3 {(0)}^{2} - 25 = 0 - 25 = - 25$ $y=f(-b/(2a))=f(0)=3(0)^2-25=0-25=-25$

$V e r t e x \Rightarrow (x, y) \Rightarrow (0, - 25)$ $Vertex => (x,y) => (0,-25)$

Click on the video below to learn about several methods of finding the vertex of a quadratic equation.

How do you find the vertex of f(x)=3x2−25f(x)=3x^2-25?