evanrosario11
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Find the values of a, b, and c for the equation in standard form (y=ax^2+bx+c) of the graph of the parabola shown below.

a =____________
b = ___________
c = ___________

Find The Values Of A B And C For The Equation In Standard Form Yax2bxc Of The Graph Of The Parabola Shown Below A B C class=

Sagot :

Nayefx

Answer:

[tex] \boxed{ \boxed{ \mathfrak{a = - \frac{1}{2 } \: b = 2 \: c = 4}}}[/tex]

Step-by-step explanation:

to understand this

you need to know about:

  • quadratic equation
  • PEMDAS

tips and formulas:

  • the parabola crosses f(x) at (4,0)
  • the vertex of the parabola (2,6)
  • vertex:(h,k)
  • h=-b/2a
  • k=f(x)
  • vertex form:f(x)=a(x-h)²+k
  • standard form:ax²+bx+c

let's solve:

the vertex form of the equation is

  • [tex] \sf f(x) = a(x - 2 {)}^{2} + 6[/tex]

let's figure out a

since the parabola crosses f(x) at (0,4)

therefore

  • [tex] \sf a(0 - {2)}^{2} + 6 = 4[/tex]
  1. [tex] \sf simplify \: parentheses : \\ \sf a( - 2 {)}^{2} + 6 = 0[/tex]
  2. [tex] \sf simplify \: squres : \\ \sf \: 4a + 6 = 4[/tex]
  3. [tex] \sf cancel \: 6 \: from \: both \: sides : \\ \sf 4a + 6 - 6 = 4 - 6 \\ \sf 4a = - 2 [/tex]
  4. [tex] \sf divide \: both \: sides \: by \: 4 : \\ \sf \frac{4a}{4} = - \frac{ 2}{4} \\ \sf a = - \frac{1}{2} [/tex]

let's figure out b,c

  • we have to substitute the value of a into the vertex form and then simply it to get b and c
  1. [tex] \sf substitue \: the \: value \: of \: a \: into \: the \: vertex \: form \: of \: the \: equation : \\ \sf \frac{1}{2} (x - 2 {)}^{2} + 6[/tex]
  2. [tex] \sf use \: (a - b {)}^{2} = {a}^{2} - 2ab + {b}^{2} \: to \: simplify \: the \: exponet : \\ \sf - \frac{1}{2} ( {x}^{2} - 2.x.2 + {2}^{2} ) + 6 \\ \sf - \frac{1}{2} ( {x}^{2} - 4x + 4) + 6[/tex]
  3. [tex] \sf distribut e \: - \frac{1}{2} \: between \: the \: parentheses : \\ \sf - \frac{1}{2} x + 2x - 2 + 6[/tex]
  4. [tex] \sf add : \\ \sf - \frac{1}{2} x + 2x + 4[/tex]

therefore

  • [tex] \sf \: b = 2[/tex]
  • [tex] \sf \: c = 4[/tex
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