How do you find the vertex and the intercepts for $f (x) = - x^{2} + 2 x + 5$ $f(x) = -x^2 + 2x + 5$ ?

Question

How do you find the vertex and the intercepts for $f (x) = - x^{2} + 2 x + 5$ $f(x) = -x^2 + 2x + 5$ ?

2 Answers

Impact of this question

9711 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-06T07:48:37

it vertex is maximum at $(1, 6)$ $(1,6)$

x-intercepts $= 1 \pm \sqrt{6}$ $= 1 +- sqrt6$

y-intercept $= 5$ $= 5$

Explanation:

$f (x) = - x^{2} + 2 x + 5$ $f(x) = -x^2 + 2x + 5$

$f (x) = - (x^{2} - 2 x) + 5$ $f(x) = -(x^2 -2x) + 5$

$f (x) = - {(x - 1)}^{2} + {(- 1)}^{2} + 5$ $f(x) = -(x -1)^2 +(-1)^2 + 5$

$f (x) = - {(x - 1)}^{2} + 6$ $f(x) = -(x -1)^2 +6$

since coefficient of $(x - 1)$ $(x-1)$ is -ve value, it vertex is maximum at $(1, 6)$ $(1,6)$

to find x-intercept, plug in $f (x) = 0$ $f(x) =0$ in the equation, therefore
$0 = - {(x - 1)}^{2} + 6$ $0 = -(x -1)^2 +6$
${(x - 1)}^{2} = 6$ $(x -1)^2 = 6$
$x - 1 = \pm \sqrt{6}$ $x -1 = +- sqrt6$
$x = 1 \pm \sqrt{6}$ $x = 1 +- sqrt6$

to find y-intercept, plug in $x = 0$ $x =0$ in the equation, therefore
$f (0) = - {(0 - 1)}^{2} + 6$ $f(0) = -(0-1)^2 +6$
$f (0) = - {(- 1)}^{2} + 6$ $f(0) = -( -1)^2 +6$
$f (0) = - 1 + 6 = 5$ $f(0) = -1 +6 = 5$

Answer 2 · 2017-06-06T10:05:12

$see explanation$ $"see explanation"$

Explanation:

$for the standard form of a parabola y = a x^{2} + b x + c$ $"for the standard form of a parabola " y=ax^2+bx+c$

$then x_{vertex} = - \frac{b}{2 a}$ $"then " x_(color(red)"vertex")=-b/(2a)$

$y = - x^{2} + 2 x + 5 is in this form$ $y=-x^2+2x+5" is in this form"$

$with a = - 1, b = 2 and c = 5$ $"with " a=-1,b=2" and " c=5$

$\Rightarrow x_{vertex} = - \frac{2}{- 2} = 1$ $rArrx_(color(red)"vertex")=-2/(-2)=1$

$substitute this value into function for y-coordinate$ $"substitute this value into function for y-coordinate"$

$\Rightarrow y_{vertex} = - {(1)}^{2} + (2 \times 1) + 5 = 6$ $rArry_(color(red)"vertex")=-(1)^2+(2xx1)+5=6$

$\Rightarrow vertex = (1, 6)$ $rArrcolor(magenta)"vertex "=(1,6)$

$for intercepts$ $color(blue)" for intercepts"$

$∙ let x = 0, in function for y-intercept$ $• " let x = 0, in function for y-intercept"$

$∙ let y = 0, in function for x-intercepts$ $• " let y = 0, in function for x-intercepts"$

$x = 0 \to y = 0 + 0 + 5 = 5 \leftarrow y-intercept$ $x=0toy=0+0+5=5larrcolor(red)" y-intercept"$

$y = 0 \to - x^{2} + 2 x + 5 = 0$ $y=0to-x^2+2x+5=0$

$solve using the quadratic formula$ $"solve using the "color(blue)"quadratic formula"$

$x = \frac{- 2 \pm \sqrt{4 + 20}}{- 2} = \frac{- 2 \pm \sqrt{24}}{- 2} = \frac{- 2 \pm 2 \sqrt{6}}{- 2}$ $x=(-2+-sqrt(4+20))/(-2)=(-2+-sqrt24)/-2=(-2+-2sqrt6)/-2$

$\Rightarrow x = 1 \pm \sqrt{6}$ $rArrx=1+-sqrt6$

$\Rightarrow x \approx 3.45, x \approx - 1.45 \leftarrow x-intercepts$ $rArrx~~3.45,x~~-1.45larrcolor(red)" x-intercepts"$
graph{-x^2+2x+5 [-12.65, 12.66, -6.34, 6.3]}

Science

Math

Humanities

... and beyond

How do you find the vertex and the intercepts for $f (x) = - x^{2} + 2 x + 5$ $f(x) = -x^2 + 2x + 5$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the vertex and the intercepts for f(x)=−x2+2x+5f(x) = -x^2 + 2x + 5?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you find the vertex and the intercepts for $f (x) = - x^{2} + 2 x + 5$ $f(x) = -x^2 + 2x + 5$ ?