Question

NO LINKS!! Please help me with this statement Part 2mm​

Answer 1

`f(x)=2x^2+8x-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:

• Vertex = (-2, -13)
• Point = (0, -5)

Substitute the given vertex and point into the Vertex formula and solve for a:

`\implies -5=a(0-(-2))^2+(-13)`

`\implies -5=a(0+2)^2-13`

`\implies -5=4a-13`

`\implies 4a=8`

`\implies a=2`

Substitute the given vertex and found value of a into the Vertex formula:

`y=2(x+2)^2-13`

The standard form of a quadratic function is f(x) = ax² + bx + c

Expand the function in vertex form to standard form:

`\implies y=2(x^2+4x+4)-13`

`\implies y=2x^2+8x+8-13`

`\implies y=2x^2+8x-5`

Answer 2

• y = 2x² + 8x - 5

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Vertex form of a quadratic function:

• y = a(x - h)² + k, where (h, k) is vertex and a - constant

Given (h, k) = (-2, -13) and a point (0, - 5).

Substitute all into equation and solve for a:

• -5 = a(0 - (-2))² - 13
• -5 = 4a - 13
• 4a = 13 - 5
• 4a = 8
• a = 2

The parabola is:

• y = 2(x + 2)² - 13

Convert it to the standard form:

• y = 2(x + 2)² - 13
• y = 2(x² + 4x + 4) - 13
• y = 2x² + 8x + 8 - 13
• y = 2x² + 8x - 5