Question

I'm having a problem grabbing and figuring out quadratic equation I went to the picture

Answer 1

To obtain the vertex of the function, compare the given equation to the following equation.

`f(x)=ax^2+bx+c`

From the given, we obtain the following:

`\begin{gathered} a=-1 \\ b=8 \\ c=-10 \end{gathered}`

Substitute the obtained values to the vertex (h,k), where the values of h and k are as follows:

`\begin{gathered} h=-(b)/(2a) \\ =-(8)/(2(-1)) \\ =-(8)/(-2) \\ =-\mleft(-4\mright) \\ =4 \\ \\ k=f\mleft(h\mright) \\ =f\mleft(4\mright) \\ =-4^2+8\mleft(4\mright)-10 \\ =-16+32-10 \\ =6 \end{gathered}`

Thus, the vertex is at (4,6).

Since the variable x is raised to 2 and the value of a is negative, the parabola opens downwards.

Since the parabola opens downwards, the equation of symmetry is as follows.

`\begin{gathered} x=h \\ x=4 \end{gathered}`

Since the parabola opens downwards, the parabola has a maximum point at its vertex. Thus, f has a maximum of 6.

To obtain the x-intercepts, substitute 0 for f(x) and then solve for the value of x.

`\begin{gathered} 0=-x^2+8x-10 \\ x^2-8x+10=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(1)(10)}}{2(1)} \\ x=\frac{8\pm\sqrt[]{64-40}}{2} \\ x=\frac{8\pm\sqrt[]{24}}{2} \\ x=\frac{8\pm2\sqrt[]{6}}{2} \\ x=4\pm\sqrt[]{6} \\ \\ x_1=4+\sqrt[]{6}\approx6.45 \\ x_2=4-\sqrt[]{6}\approx1.55 \end{gathered}`

Thus, the x-intercepts are (1.55,0) and (6.45,0).

To obtain the y-intercept, substitute 0 for x and y for f(x). Then, solve for y.

`\begin{gathered} y=-x^2+8x-10 \\ y=-0^2+8(0)-10 \\ y=0+0-10 \\ y=-10 \end{gathered}`

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

To graph the equation, plot the vertex and the intercepts. Draw a smooth curve passing through the points from left to right.

Therefore, the graph of the quadratic function is shown below.