Chapter 2, Problem 11CR

To determine

Towrite:The standard form of the quadratic function whose graph is a parabola with the indicated vertex and that passes through the indicated point.

Answer to Problem 11CR

The standard form of the quadratic function that represents a parabola with vertex $\left(-3,4\right)$ and that passes through point $\left(0,-5\right)$ is $y=-{\left(x+3\right)}^2+4$ .

Explanation of Solution

Given information:

The vertex of parabola is $\left(-3,4\right)$ and passes through point $\left(0,-5\right)$

Formulaused:

The standard form of a quadratic function that represents a parabola is $y=a{\left(x-h\right)}^2+k$ . The vertex of parabola is $\left(h,k\right)$ .

Calculations:

Let thestandard form of the quadratic function that represents a parabola with vertex $\left(-3,4\right)$ and that passes through point $\left(0,-5\right)$ is

  $y=a{\left(x-h\right)}^2+k$........(1)

Therefore, we have $\left(h,k\right)\equiv \left(-3,4\right)$

  $\Rightarrow h=-3,k=4$ .

Substituting values of h, k in equation (1)

  $y=a{\left(x-(-3)\right)}^2+4$

  $y=a{\left(x+3\right)}^2+4$........(2)

Equation (2) represents the parabola passing through $\left(0,-5\right)$ , so point satisfy the equation, i.e.,

  $-5=a{\left(0+3\right)}^2+4$

  $\Rightarrow -5=9a+4$

  $\Rightarrow 9a=-9$

  $\Rightarrow a=\frac{-9}{9}=-1$

Thus, the standard form of the quadratic function that represents a parabola with vertex $\left(-3,4\right)$ and that passes through point $\left(0,-5\right)$ is $y=-1\cdot {\left(x+3\right)}^2+4$ .

Conclusion:

The standard form of the quadratic function that represents a parabola with vertex $\left(-3,4\right)$ and that passes through point $\left(0,-5\right)$ is $y=-{\left(x+3\right)}^2+4$ .