91Ó°ÊÓ

Consider the quadratic function $f\left(x\right)=-2x^2+8x-10$.

For this parabola, the graph will open downwards as the leading coefficient is negative.

Then finding the axis of symmetry, vertex, the intercepts, and a point with help of the axis of symmetry, will help us to plot the points of the graph and get the required graph.

On comparing the given function $f\left(x\right)=-2x^2+8x-10$with the standard form $f\left(x\right)=a{x}^2+bx+c$we get

$$\begin{gathered} a=-2 \\ b=8 \\ c=-10 \end{gathered}$$

The axis of symmetry for the standard form is $x=-\frac{b}{2a}$.

So for the given function, the axis of symmetry is given as

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{8}{2·(-2)} \\ x=-(-2) \\ x=2 \end{gathered}$$

Thus the equation of the axis of symmetry is $x=2$.

Vertex is a point on the line of axis of symmetry.

Here axis of symmetry is $x=2$.

So the function value $f\left(2\right)$is given as

$$\begin{gathered} f\left(2\right)=-2·2^2+8·2-10 \\ f\left(2\right)=-8+16-10 \\ f\left(2\right)=-2 \end{gathered}$$

So the vertex of the parabola is $(2,-2)$.

To find the y-intercept substitute $0$for $x$in the function.

$$\begin{gathered} f\left(0\right)=-2·0^2+8·0-10 \\ f\left(0\right)=-0+0-10 \\ f\left(0\right)=-10 \end{gathered}$$

Thus the y intercept is $(0,-10)$.

To find $x$intercepts substitute $0$for $f\left(x\right)$in the function.

$\begin{array}{rcl}f\left(x\right)& =& -2x^2+8x-10\\ 0& =& -2x^2+8x-10\end{array}$

The discriminant of the function is given as

$$\begin{gathered} D=8^2-4(-2)(-10) \\ D=64-80 \\ D=-16 \end{gathered}$$

As the value of the discriminant is less than zero, so the equation has no real roots.

And so the function has no $x$intercepts.

We have y intercept as $(0,-10)$.

The axis of symmetry is$(2,-2)$.
It means that the intercept is at a distance of two units to the left of the axis of symmetry.
We can get a point on the graph using this information.
A point at the same distance, i.e., two units, but to the right will also lie on the graph. This point will be $(4,-10)$.

Plotting all the points we got and joining these points will give us the graph as-