Question

Draw a sketch of the quadratic function

y = - x^{2} + 2 x - 4

using vertex, axis of symmetry, concavity,

x and y

intercepts.

Accepted Answer

Given quadratic function is $y = - x^{2} + 2 x - 4$

Comparing the above quadratic function with $y = a x^{2} + b x + c$

We have $a = - 1, b = 2, c = - 4$

The coordinates of the vertex of quadratic function $y = a x^{2} + b x + c$ are $(- \frac{b}{2 a}, f (- \frac{b}{2 a}))$

Now, $- \frac{b}{2 a} = - \frac{2}{2 (- 1)} = 1$ and $f (- \frac{b}{2 a}) = f (1) = - 1^{2} + 2 (1) - 4 = - 3$

Therefore, the coordinates of the vertex of given quadratic function are $(1, - 3)$ .

The equation of axis of symmetry of the quadratic function $y = a x^{2} + b x + c$ is $x = - \frac{b}{2 a}$

$\Rightarrow x = - \frac{2}{2 (- 1)}$

$\Rightarrow x = 1$ .

Here $a < 0$ , so the quadratic function $y = - x^{2} + 2 x - 4$ is concave down.

Therefore, $y -$ intercept of the quadratic function is $(0, - 4)$ .

We know that a quadratic function is factorizable if discriminant i.e., $D = b^{2} - 4 a c$ is either $0$ or a positive perfect square.

Now, $D = {(2)}^{2} - 4 (- 1) (- 4) = 4 + 16 = 20$ (which is neither $0$ nor a positive perfect square)

So, given function is not factorizable

From the above results the graph of the quadratic function is

Rose Harrison and Clara Huizink Solutions for Exercise 8: Practice 1

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison and Clara Huizink Solutions for Exercise 8: Practice 1

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