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Given a parabola with the vertex (2, 5), and a point (-3, 1), choose the equation that would solve for "a" to be able to write the vertex form of the graph.
01 a(-2-3)² +5
01= a(-3-2)² - 5
01= a(-3+2)2² +5
01= a(-3-2)² +5



Answer :

mhanifa

Answer:

  • D) 1 = a( - 3 - 2)² + 5

Step-by-step explanation:

The vertex form is:

  • f(x) = a(x - h)² + k, where (h, k) is the vertex

Given

  • (h, k) = (2, 5) and a point (-3, 1)

Substitute into given form

  • f(x) = 1, x = - 3, h = 2, k = 5

  • 1 = a( - 3 - 2)² + 5

Correct choice is D

semsee45

Answer:

[tex]\textsf{d)} \quad 1=a(-3-2)^2+5[/tex]

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{5.6 cm}\underline{Vertex form of a quadratic equation}\\\\$y=a(x-h)^2+k$\\\\where:\\ \phantom{ww}$\bullet$ $(h,k)$ is the vertex. \\ \phantom{ww}$\bullet$ $a$ is some constant.\\\end{minipage}}[/tex]

Given information:

  • Vertex = (2, 5)
  • Point on the parabola = (-3, 1)

Therefore:

  • h = 2
  • k = 5
  • x = -3
  • y = 1

To find the constant "a", substitute found values into the vertex formula:

[tex]\implies 1=a(-3-2)^2+5[/tex]

Additional information

Solve the equation for a:

[tex]\implies 1=a(-5)^2+5[/tex]

[tex]\implies 1=25a+5[/tex]

[tex]\implies -4=25a[/tex]

[tex]\implies a=-\dfrac{4}{25}[/tex]

Therefore, the equation of the parabola in vertex form is:

[tex]y=-\dfrac{4}{25}\left(x-2\right)^2+5[/tex]

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