Question

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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 1

• D) 1 = a( - 3 - 2)² + 5

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

Answer 2

`\textsf{d)} \quad 1=a(-3-2)^2+5`

Explanation:

`\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}}`

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:

`\implies 1=a(-3-2)^2+5`

Additional information

Solve the equation for a:

`\implies 1=a(-5)^2+5`

`\implies 1=25a+5`

`\implies -4=25a`

`\implies a=-(4)/(25)`

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

`y=-(4)/(25)\left(x-2\right)^2+5`