Question

Sketch the graph for the following quadratic function.

Answer 1

Answer:

Please refer to the attached image for the graph of given function.

Step-by-step explanation:

Given the equation:

$-x^{2} +4x+12$

Let us rewrite by letting it equal to $y$.

$y=-x^{2} +4x+12$

Now, we can see that it is a quadratic equation and it is known that a quadratic equation has a graph of parabola.

Let us compare the given equation with standard quadratic equation:

$y=ax^{2} +bx+c$

we get:

$a = -1\\b = 4\\c = 12$

Coefficient of $x^{2}$ is negative 1, so the parabola will open downwards.

Axis of symmetry: It is the line which will divide the parabola in two equal congruent halves.

Formula for axis of symmetry is:

$x = -\dfrac{b}{2a}$

$x = -\dfrac{4}{2(-1)}\\\Rightarrow x=2$

It is shown as dotted line in the image attached in the answer area.

Axis of symmetry will also contain the vertex of the parabola.

It is a downward parabola so vertex will be the highest point on this parabola.

Putting x = 2 in the equation of parabola:

$y=-2^{2} +4\times 2+12\\\Rightarrow y =16$

So, vertex will be at P(2, 16).

Now, let us find points of parabola to sketch graph:

put x = 0, $y=-0^{2} +4\times 0+12=12$

Another point is Y(0,12)

Now, let us put y = 0, it will give us two points because the equation is quadratic in x.

$0=-x^{2} +4x+12\\\Rightarrow -x^{2} +6x-2x+12=0\\\Rightarrow -x(x -6)-2(x-6)=0\\\Rightarrow (-x-2)(x-6)=0\\\Rightarrow x = -2, 6$

So, other two points are X1(-2, 0) and X2(6,0).

If we plot the points P, Y, X1 and X2 we get a graph as attached in the image in answer area.