Question
The linear function y=1/2x−5 is graphed on the same axis as the quadratic function. Enter the coordinates of the solution with a positive x-coordinate. (1 point) The coordinates of the solution with a positive x-value are (,).
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Answered by
GPT-4o mini
To find the coordinates of the solution where the linear function \( y = \frac{1}{2}x - 5 \) intersects with a quadratic function, we first need to know the equation of the quadratic function. Since you didn't provide the equation, I'll represent it generally as \( y = ax^2 + bx + c \).
To determine the points of intersection, we set the two equations equal to each other:
\[
\frac{1}{2}x - 5 = ax^2 + bx + c
\]
Rearranging this equation leads us to:
\[
ax^2 + (b - \frac{1}{2})x + (c + 5) = 0
\]
You would then solve this quadratic equation for \( x \) using the quadratic formula:
\[
x = \frac{-(b - \frac{1}{2}) \pm \sqrt{(b - \frac{1}{2})^2 - 4a(c + 5)}}{2a}
\]
From the solutions for \( x \), you would identify the one with the positive value.
If you provide the specific quadratic function, I can give you a more exact answer or calculate it directly.
To determine the points of intersection, we set the two equations equal to each other:
\[
\frac{1}{2}x - 5 = ax^2 + bx + c
\]
Rearranging this equation leads us to:
\[
ax^2 + (b - \frac{1}{2})x + (c + 5) = 0
\]
You would then solve this quadratic equation for \( x \) using the quadratic formula:
\[
x = \frac{-(b - \frac{1}{2}) \pm \sqrt{(b - \frac{1}{2})^2 - 4a(c + 5)}}{2a}
\]
From the solutions for \( x \), you would identify the one with the positive value.
If you provide the specific quadratic function, I can give you a more exact answer or calculate it directly.
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