What is the vertex of $y = x^{2} + 3 x + 5$ $y=x^2+3x+5$ ?

Question

What is the vertex of $y = x^{2} + 3 x + 5$ $y=x^2+3x+5$ ?

1 Answer

Impact of this question

2383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-10T12:32:16

$V e r t e x (- \frac{3}{2}, \frac{11}{4})$ $Vertex (-3/2, 11/4)$

Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{3}{2}$ $x = -b/(2a) = -3/2$
y-coordinate of vertex:
$y (- \frac{3}{2}) = (\frac{9}{4}) + 3 (- \frac{3}{2}) + 5 = \frac{9}{4} - \frac{9}{2} + 5 = \frac{11}{4}$ $y(-3/2) = (9/4) + 3(-3/2) + 5 = 9/4 - 9/2 + 5 = 11/4$

$V e r t e x (- \frac{3}{2}, \frac{11}{4})$ $Vertex (-3/2, 11/4)$

Science

Math

Humanities

... and beyond

What is the vertex of $y = x^{2} + 3 x + 5$ $y=x^2+3x+5$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the vertex of y=x^2+3x+5 y=x2+3x+5y=x^2+3x+5 ?

1 Answer

Explanation:

Related questions

Impact of this question

What is the vertex of $y = x^{2} + 3 x + 5$ $y=x^2+3x+5$ ?