Answer:
i) y = 2(x + 2)² - 5
ii) y = x² + 2x + 6
iii) y = (x - 3)(x - 5)
iv) y = x(x - 4)
Step-by-step explanation:
Question i
To write an equation for a quadratic function that has a vertex at (-2, -5), we can use the vertex form of a quadratic function:
\(\boxed{\begin{array}{l}\underline{\textsf{Vertex form of a quadratic equation}}\\\\y=a(x-h)^2+k\\\\\textsf{where:}\\\phantom{ww}\bullet\;(h,k)\;\sf is\;the\;vertex.\\\phantom{ww}\bullet\;a\;\sf is\;the\;leading\;coefficient.\\\end{array}}\)
As (h, k) is the vertex, then:
The leading coefficient (a) does not affect the vertex, so let's choose a = 2:
\(y=2(x-(-2))^2 - 5\\\\y=2(x+2)^2 - 5\)
Therefore, an equation for a quadratic function that has a vertex at (-2, -5) is:
\(\Large\boxed{\boxed{y=2(x+2)^2 - 5}}\)
\(\dotfill\)
Question ii
To write an equation for a quadratic function that has a y-intercept at (0, 6), we can use the standard form of a quadratic function:
\(\boxed{\begin{array}{l}\underline{\textsf{Standard form of a quadratic equation}}\\\\y=ax^2+bx+c\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a$ is the leading coefficient}.\\\phantom{ww}\bullet\;\textsf{$(0, c)$ is the $y$-intercept}.\end{array}}\)
Given that (0, 6) is the y-intercept, then c = 6.
We can use any integer for 'a' and 'b'.
Therefore, an equation for a quadratic function that has a y-intercept at (0, 6) is:
\(\Large\boxed{\boxed{y=x^2+2x+6}}\)
\(\dotfill\)
Question iii
To write an equation for a quadratic function that has x-intercepts at (3, 0) and (5, 0), we can use the factored form of a quadratic function:
\(\boxed{\begin{array}{l}\underline{\textsf{Factored form of a quadratic equation}}\\\\y=a(x-r_1)(x-r_2)\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a$ is the leading coefficient}.\\\phantom{ww}\bullet\;\textsf{$r_1$ and $r_2$ are the $x$-intercepts (roots)}.\end{array}}\)
Given that the x-intercepts are at (3, 0) and (5, 0), then:
Substitute these into the equation, using any integer for 'a':
\(y=1(x-3)(x-5)\\\\y=(x-3)(x-5)\)
Therefore, an equation for a quadratic function that has x-intercepts at (3, 0) and (5, 0) is:
\(\Large\boxed{\boxed{y=(x-3)(x-5)}}\)
\(\dotfill\)
Question iv
To write an equation for a quadratic function that has x-intercepts at the origin (0, 0) and (4, 0), we can use the factored form of a quadratic function:
\(\boxed{\begin{array}{l}\underline{\textsf{Factored form of a quadratic equation}}\\\\y=a(x-r_1)(x-r_2)\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a$ is the leading coefficient}.\\\phantom{ww}\bullet\;\textsf{$r_1$ and $r_2$ are the $x$-intercepts (roots)}.\end{array}}\)
Given that the x-intercepts are at (0, 0) and (4, 0), then:
Substitute these into the equation, using any integer for 'a':
\(y=1(x-0)(x-4)\\\\y=x(x-4)\)
Therefore, an equation for a quadratic function that has x-intercepts at (0, 0) and (4, 0) is:
\(\Large\boxed{\boxed{y=x(x-4)}}\)
