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consider the graph quadratic function with one point located at Point p. plot a point on the graph that is integer coordinates and represents an average rate of change of 5 with Point p

consider the graph quadratic function with one point located at Point p. plot a point-example-1

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In order to find the point, let's find the equation of a line with slope 5 and that passes through the point P.

To do so, let's use the slope-intercept form y = mx + b with the slope m = 5, and then use the point (x, y) = (1, 8):


\begin{gathered} y=5x+b\\ \\ 8=5+b\\ \\ b=3 \end{gathered}

Now, to find the required point, we need to find the second point that is on the line and on the parabola.

To do so, let's find the equation of the parabola, with vertex at (2, 9) and y-intercept at (0, 5). Also, let's use the x-intercept (5, 0):


\begin{gathered} y=ax^2+bx+c\\ \\ (0,5):\\ \\ 5=0+0+c\\ \\ c=5\\ \\ \\ \\ (2,9):\\ \\ 9=4a+2b+5\\ \\ 4a+2b=4\\ \\ 2a+b=2 \\ \\ \\ (5,0):\\ \\ 0=25a+5b+5\\ \\ 25a+5b=-5\\ \\ 5a+b=-1 \end{gathered}

Solving this system, we have a = -1 and b = 4, so the parabola is y = -x² + 4x + 5.

Now, let's equate the equation of the parabola with the equation of the line:


\begin{gathered} -x^2+4x+5=5x+3\\ \\ -x^2-x+2=0\\ \\ x^2+x-2=0\\ \\ x=(-b\pm√(b^2-4ac))/(2a)\\ \\ x=(-1\pm√(1+8))/(2)\\ \\ x=(-1\pm3)/(2)\\ \\ x_1=(-1-3)/(2)=(-4)/(2)=-2\\ \\ x_2=(-1+3)/(2)=(2)/(2)=1 \end{gathered}

Therefore the second point is located at x = -2. The y-value is:


\begin{gathered} y=5x+3\\ \\ y=5\cdot(-2)+3\\ \\ y=-10+3\\ \\ y=-7 \end{gathered}

So the answer is the point at (-2, -7).

User Haseeb Hassan Asif
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