Question

NO LINKS!!!
Write the standard form of the quadratic function whose graph is a parabola with the given vertex and that passes through the given point. (Let x be the independent variable and y be the dependent variable.)

Answer 1

Answer:

$y=3x^2-6x-2$

Step-by-step 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 = (1, -5)
• Point = (-1, 7)

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

$\implies 7=a(-1-1)^2-5$

$\implies 7=a(-2)^2-5$

$\implies 7=4a-5$

$\implies 12=4a$

$\implies a=3$

Substitute the found value of a together with the vertex into the formula and expand to standard form:

$\implies y=3(x-1)^2-5$

$\implies y=3(x-1)(x-1)-5$

$\implies y=3(x^2-2x+1)-5$

$\implies y=3x^2-6x+3-5$

$\implies y=3x^2-6x-2$