How do you find the VERTEX of a parabola $f (x) = - x^{2} + 3 x + 10$ $f(x)= -x^2 + 3x + 10$ ?

Question

2853 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-17T22:12:17

Find vertex of $f (x) = - x^{2} + 3 x + 10$ $f(x) = -x^2 + 3x + 10$
Vertex $(\frac{3}{2}, \frac{49}{4})$ $(3/2, 49/4)$

x coordinate of vertex: $x = (- \frac{b}{2} a) = - \frac{3}{- 2} = \frac{3}{2}$ $x = (-b/2a) = -3/-2 = 3/2$

y coordinate of vertex: $y = f (\frac{3}{2}) = - \frac{9}{4} + \frac{9}{2} + 10 = \frac{49}{4}$ $y = f(3/2) = - 9/4 + 9/2 + 10 = 49/4$

Answer 2 · 2015-07-17T22:31:47

$x = \frac{3}{2} or 1 \frac{1}{2}$ $x=3/2 or 1 1/2$

$y = \frac{49}{4} or 12 \frac{1}{4}$ $y=49/4 or 12 1/4$

The vertex is $(\frac{3}{2}, \frac{49}{4})$ $(3/2, 49/4)$ or $(1 \frac{1}{2}, 12 \frac{1}{4})$ $(1 1/2, 12 1/4)$ .

$f (x) = - x^{2} + 3 x + 10$ $f(x)=-x^2+3x+10$

Substitute $y$ $y$ for $f (x)$ $f(x)$ .

$y = - x^{2} + 3 x + 10$ $y=-x^2+3x+10$ is in the form $a x^{2} + b x + c$ $ax^2+bx+c$ , where $a = - 1, b = 3, c = 10$ $a=-1, b=3, c=10$ .

To find the value of $x$ $x$ , use the equation $x = \frac{- b}{2 a}$ $x=(-b)/(2a)$ .

$x = \frac{- 3}{2 \cdot - 1}$ $x=(-3)/(2*-1)$

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

$x = \frac{3}{2} = 1 \frac{1}{2}$ $x=3/2=1 1/2$

To find the $y$ $y$ value, substitute $\frac{3}{2}$ $3/2$ for $x$ $x$ in the equation $y = x^{2} + 3 x + 10$ $y=x^2+3x+10$ .

$y = - {(\frac{3}{2})}^{2} + 3 (\frac{3}{2}) + 10$ $y=-(3/2)^2+3(3/2)+10$ =

$y = - \frac{9}{4} + \frac{9}{2} + 10$ $y=-9/4+9/2+10$

The common denominator is $4$ $4$ . Multiply the terms on the right side by the fraction that will give each term a denominator of $4$ $4$ .

$y = - \frac{9}{4} + \frac{9}{2} \cdot \frac{2}{2} + 10 \cdot \frac{4}{4}$ $y=-9/4+9/2*2/2+10*4/4$

$y = - \frac{9}{4} + \frac{18}{4} + \frac{40}{4}$ $y=-9/4+18/4+40/4$ =

$y = \frac{49}{4} = 12 \frac{1}{4}$ $y=49/4=12 1/4$ .

How do you find the VERTEX of a parabola f(x)=−x2+3x+10f(x)= -x^2 + 3x + 10?