Question

q. write the equation of the graph in factored form.

Explanation:

Step1: Identify the roots of the parabola

From the graph, the parabola intersects the x - axis at \(x = 10\) (a double root? Wait, no, looking at the graph, the parabola seems to open to the left (since it's a horizontal parabola). Wait, the standard form of a horizontal parabola is \((y - k)^2=4p(x - h)\) or factored form for a quadratic in \(y\) or \(x\). Wait, actually, if it's a parabola opening to the left, the equation can be written in factored form related to its roots. Wait, let's check the intercepts. The graph intersects the x - axis at \(x = 10\) (wait, no, the x - axis is the horizontal line \(y = 0\)). Wait, looking at the grid, the vertex? Wait, maybe it's a parabola that is a function of \(y\), so \(x=ay^{2}+by + c\). Let's find the roots. When \(y = 0\), \(x = 10\)? Wait, no, the graph passes through \((10,0)\) and maybe another point? Wait, no, the parabola is opening to the left, so it's a horizontal parabola. Let's assume it's a parabola with vertex at \((h,k)\) and opening left. The general factored form for a horizontal parabola (quadratic in \(y\)) is \(x=a(y - y_1)(y - y_2)\), where \(y_1\) and \(y_2\) are the \(y\) - intercepts (when \(x = 0\)). Wait, when \(x = 0\), let's find the \(y\) - values. From the graph, when \(x = 0\), the parabola passes through \(y = 5\) and \(y=-5\)? Wait, no, looking at the grid, the vertical axis (y - axis) has marks. Wait, maybe the roots (the points where \(x = 0\)) are \(y = 5\) and \(y=-5\), and it passes through \((10,0)\)? Wait, no, when \(y = 0\), \(x = 10\)? Wait, the graph at \(y = 0\) (the x - axis) touches or crosses at \(x = 10\)? Wait, no, the blue curve: when \(y = 0\), \(x = 10\), and when \(x = 0\), \(y = 5\) and \(y=-5\)? Wait, let's re - examine. The graph is a horizontal parabola (opening to the left) with \(x\) - intercept at \((10,0)\) and \(y\) - intercepts at \((0,5)\) and \((0, - 5)\). So the equation in factored form (since it's a quadratic in \(y\)) is \(x=a(y - 5)(y + 5)\). Now, we can find \(a\) by plugging in the point \((10,0)\). Wait, no, when \(y = 0\), \(x=a(0 - 5)(0 + 5)=a(-5)(5)=-25a\). We know that when \(y = 0\), \(x = 10\)? Wait, no, that can't be. Wait, maybe I got the intercepts wrong. Wait, the graph: the left - most part? No, the blue curve is a parabola opening to the left, so the vertex is at \((h,k)\), and it has two \(y\) - intercepts (when \(x = 0\)) and one \(x\) - intercept? Wait, no, a horizontal parabola (quadratic in \(y\)) has two \(y\) - intercepts (when \(x = 0\)) and is symmetric about the x - axis (if the vertex is on the x - axis). Wait, looking at the graph, it's symmetric about the x - axis (the horizontal line \(y = 0\)). So the vertex is on the x - axis, at \((h,0)\). The general equation of a horizontal parabola symmetric about the x - axis is \(x=a(y - 0)^2+h\), or in factored form, since it's a quadratic in \(y\), \(x=a y^{2}+h\). But we need factored form. Wait, if it has roots at \(y = 5\) and \(y=-5\) (when \(x = 0\)), then \(x=a(y - 5)(y + 5)\). Now, let's find a point on the parabola. Let's take the point \((0,5)\): plug \(x = 0\), \(y = 5\) into \(x=a(y - 5)(y + 5)\), we get \(0=a(5 - 5)(5 + 5)=0\), which is not helpful. Wait, take the point \((10,0)\): plug \(x = 10\), \(y = 0\) into \(x=a(y - 5)(y + 5)\), we get \(10=a(0 - 5)(0 + 5)=a(-25)\), so \(a=\frac{10}{-25}=-\frac{2}{5}\). Wait, no, that would give \(x=-\frac{2}{5}(y - 5)(y + 5)\). Let's check when \(y = 0\), \(x=-\frac{2}{5}(-5)(5)=-\frac{2}{5}\times(-25)=10\), which matches. When \(x = 0\), \(0=-\frac{2}{5}(y - 5)(y + 5)\), so \((y - 5)(y + 5)=0\), so \(y = 5\) or \(y=-5\), which also matches. But the problem says "factored form". Wait, maybe it's a vertical parabola? No, it's opening to the left, so it's a horizontal parabola. Wait, but maybe the equation is \(x=-\frac{1}{5}(y - 5)(y + 5)\)? Wait, no, let's recalculate. Wait, if the parabola passes through \((10,0)\), \((0,5)\) and \((0,-5)\), then using \(x=a(y - 5)(y + 5)\), plugging \((10,0)\): \(10=a(0 - 5)(0 + 5)\Rightarrow10=-25a\Rightarrow a=-\frac{10}{25}=-\frac{2}{5}\). So \(x=-\frac{2}{5}(y - 5)(y + 5)\). But maybe we can write it as \(x=-\frac{1}{5}(y^{2}-25)\), but factored form is \(x=-\frac{2}{5}(y - 5)(y + 5)\). Wait, but maybe the graph is a parabola with vertex at \((10,0)\) and opening left, so the standard form is \((y - 0)^2=4p(x - 10)\), and it passes through \((0,5)\). Then \(25 = 4p(-10)\Rightarrow4p=-\frac{25}{10}=-\frac{5}{2}\), so the equation is \(y^{2}=-\frac{5}{2}(x - 10)\), which can be rewritten as \(x=-\frac{2}{5}y^{2}+10\). To write in factored form (quadratic in \(y\)), \(x=-\frac{2}{5}(y^{2}-25)=-\frac{2}{5}(y - 5)(y + 5)\).

Wait, maybe I made a mistake in the direction. Alternatively, if it's a vertical parabola (opening up or down), but the graph is opening to the left, so it's horizontal. So the factored form (for a horizontal parabola, quadratic in \(y\)) is \(x = a(y - y_1)(y - y_2)\), where \(y_1\) and \(y_2\) are the \(y\) - intercepts (when \(x = 0\)). From the graph, when \(x = 0\), \(y = 5\) and \(y=-5\), so \(y_1 = 5\), \(y_2=-5\), and when \(y = 0\), \(x = 10\). Plugging into \(x=a(y - 5)(y + 5)\) with \(y = 0\), \(x = 10\): \(10=a(0 - 5)(0 + 5)\Rightarrow10=-25a\Rightarrow a=-\frac{2}{5}\). So the equation is \(x=-\frac{2}{5}(y - 5)(y + 5)\). But maybe the problem expects a simpler form, maybe with \(a =-\frac{1}{5}\)? Wait, no, let's check the grid again. The x - axis (horizontal) has marks: from 0 to 10, 15, etc. The y - axis (vertical) has marks: 5, 0, - 5. So when \(y = 5\), \(x = 0\); \(y=-5\), \(x = 0\); \(y = 0\), \(x = 10\). So the three points are \((0,5)\), \((0,-5)\), \((10,0)\). So the quadratic equation in \(x\) (as a function of \(y\)) is \(x=ay^{2}+bx + c\)? No, \(x\) as a function of \(y\): \(x=ay^{2}+k\). Wait, when \(y = 0\), \(x = 10\), so \(x=ay^{2}+10\). When \(y = 5\), \(x = 0\), so \(0=a(25)+10\Rightarrow25a=-10\Rightarrow a=-\frac{10}{25}=-\frac{2}{5}\). So \(x=-\frac{2}{5}y^{2}+10=-\frac{2}{5}(y^{2}-25)=-\frac{2}{5}(y - 5)(y + 5)\).

Step2: Write the factored form

The factored form of the horizontal parabola (quadratic in \(y\)) is \(x=-\frac{2}{5}(y - 5)(y + 5)\). But maybe the problem has a simpler integer coefficient. Wait, maybe I misread the graph. Wait, maybe the parabola is \(x=-\frac{1}{5}(y - 5)(y + 5)\). Wait, no, when \(y = 0\), \(x=-\frac{1}{5}(-25)=5\), but the graph at \(y = 0\) is at \(x = 10\). So my first calculation is correct. So the factored form is \(x=-\frac{2}{5}(y - 5)(y + 5)\). But maybe the problem is a vertical parabola? No, the graph is opening to the left, so it's horizontal.

Wait, another approach: maybe the equation is \(x=-\frac{1}{5}(y^{2}-25)\), which is \(x=-\frac{1}{5}(y - 5)(y + 5)\). Wait, no, when \(y = 0\), \(x = 5\), but the graph at \(y = 0\) is at \(x = 10\). So I think the correct factored form is \(x=-\frac{2}{5}(y - 5)(y + 5)\).

Answer:

\(x =-\frac{2}{5}(y - 5)(y + 5)\)